LIBRARY 


UNIVERSITY  OF  CALIFORNIA. 


Class 


;  i  • " 


STATICS 


BY 


ALGEBRAIC   AND   GRAPHIC 
METHODS 


INTENDED  PRIMARILY  FOR  STUDENTS  OF 
ENGINEERING  AND  ARCHITECTURE 


BY 


LEWIS  J.  JOHNSON,   C.E. 


ASSISTANT    PROFESSOR    OF   CIVIL   ENGINEERING    IN    HARVARD   UNIVERSITY; 

ASSOCIATE    MEMBER    OF   THE    AMERICAN    SOCIETY 

OF  CIVIL  ENGINEERS 


FIRST   EDITION 
FIRST    THOUSAND 


NEW  YORK 

JOHN   WILEY   &   SONS 

LONDON:   CHAPMAN   &   HALL,  LIMITED 

1903. 


GENERAL 

Copyright,  1903, 


L.  J.  JOHNSON. 


ROBERT  DRUMMOND,   PRINTER,  NEW  YORK. 


PREFACE. 


THIS  volume  has  been  written  in  the  hope  of  helping 
students  of  engineering  and  architecture  to  acquire  a  knowl- 
edge of  Statics  which  will  include  the  power  to  apply  it  cor- 
rectly in  professional  work.  To  this  end  an  attempt  has  been 
made  to  carry  out  several  specific  purposes,  prominent  among 
which  may  be  mentioned  the  following : 

1 .  To  give  much  attention  to  the  starting  points  of  the  sci- 
ence, and  to  make  as  clear  as  possible  the  course  of  deduction 
therefrom. 

2.  To  point  out  the  inherent  mathematical  limitations  of 
pure    Statics,  and  to  show    how    all  its  important    problems 
are  solved. 

3.  To  develop  algebraic  and  graphic  methods  of  solution, 
or,  if  one  prefer  the  terms,  Analytical  and  Graphical  Statics, 
side  by  side  and  with  equal  thoroughness. 

4.  To  present  a  graded  set  of  problems   illustrating   not 
only  universal  principles  but  also  how  Statics  is  used  in  en^ 
gineering  practice. 

'5.  Finally,  to  keep  the  book  of  a  size  commensurate  with 
the  small  amount  of  new  matter  which  the  reader,  versed  in  the 
simplest  operations  of  elementary  mathematics,  need  master  to 
gain  the  desired  end. 

It  may  be  pointed  out  that  the  phrases  Analytical  Statics 
and  Graphical  Statics  are  avoided.  The  ground  for  this  is  that 
there  seems  to  be  no  necessity  for  using  them,  in  this  work  at 


197448 


v  PREFACE. 

least,  and  that  the  terms  seem  objectionable  as  tending  to 
obscure  the  unity  of  Statics,  and  to  produce  the  impression 
that  two  merely  alternative  methods  of  procedure  from  identi- 
cal premises  and  of  identical  mathematical  significance  are 
loosely  connected  if  not  actually  distinct  branches  of  the 
science. 

The  subject  will  be  found  developed  in  the  following  pages 
in  such  a  way  as  to  make  it  possible  to  solve  problems  from 
the  outset  by  both  methods  in  parallel  (as  in  the  plates),  and 
the  practice  of  making  such  double  solution  is  believed  to  be 
of  great  value  not  only  for  the  drill  of  checking  one's  own 
work,  but  also  for  the  clearer  light  in  which  each  method  is 
seen  by  being  kept  in  close  relation  to  the  other.  Moreover, 
as  the  student  checks  the  correctness  of  his  own  work,  it  is 
possible,  even  with  classes  of  upwards  of  one  hundred,  to 
assign  different  problem  data  to  each  student,  and  still  keep 
the  labor  of  inspecting  students'  work  at  a  minimum.  After 
some  experience  with  the  two  methods  side  by  side,  it  is 
believed  that  .practice  should  be  had  in  the  rapid  solution  of  a 
large  number  of  varied  problems  such  as  can  be  found  in  the 
familiar  works  on  Statics,  in  which  a  single  solution  by  either 
method  is  accepted  and  in  which  the  student  judges  the 
correctness  of  his  results  not  only  by  examination  of  his 
work,  but  by  reflection  upon  their  reasonableness  under  the 
conditions. 

Of  course  .many  different  sources  have  been  drawn  upon 
freely  for  suggestions,  methods,  and  material,  but  it  will  not 
be  out  of  place  to  mention  as  especially  prominent  among 
them  Rankine's  Applied  Mechanics  and  Hoskins'  Graphic 
Statics,  and  the  writer  takes  pleasure  in  acknowledging  his 
obligation  accordingly.  Grateful  recognition  is  also  due  to 
Professor  I.  N.  Hollis  for  valued  criticism. 
CAMBRIDGE,  MASS.,  July,  1902. 


CONTENTS. 


PART  I. 

GENERAL  PRINCIPLES  AND  METHODS. 
INTRODUCTION. 

SECTION  PAGE 

1.  Algebraic  and  Graphic  Methods  Compared I 

2.  History  of  Graphic  Methods  in  Statics 3 

CHAPTER  I. 

DEFINITIONS   AND  JPRELIMINARIES. 

3.  Mechanics  and  its  Subdivisions 4 

4.  Rigid  Body.     Particle   5 

5.  Rest  and  Motion. 6 

6.  Force 7 

7.  Elements  of  Force 7 

8.  Magnitude 8 

9.  Direction  and  Sense 8 

10.  Point  of  Application ....    8 

1 1.  Coplanar  and  Non-Coplanar  Forces. 9 

12.  Concurrent  and  Non-Concurrent  Forces 9 

13.  Equilibrium   9 

14.  Equivalence 10 

15.  Resultant  and  Equilibrant 10 

16.  Components   10 

17.  Relation  Between  Two  Components  and  their  Resultant.     Parallelogram 

of  Forces 10 

Exercise  1 1 1 

18.  Couples  and  their  Moments 12 

19.  Center  of  Rotation  for  a  Couple 13 

20.  Single  Force  Always  Replaceable  by  a  Single  Force  and  a  Couple 13 

21.  Moment  of  a  Force 14 

Exercise  2 15 

•  v 


VI  CONTENTS. 

SECTION  PAGE 

22.  Propositions  Regarding  the  Moment  of  a  Force    15 

23.  Proposition  Regarding  the  Moment  of  a  Couple 16 

24.  Sets  of  Forces  Classified 16 

25.  Conditions  of  Equivalence 17 

CHAPTER  II. 

NOTATION    AND   CONVENTIONS. 

26.  Notation.     Conventions  Regarding  Elements  of  Forces 19 

27.  Space  Diagram  and  Magnitude  Diagram 20 

28.  Illustration  of  Scheme  of  Notation 20 

CHAPTER  III. 

I'ARALLELOGRAM   OF   FORCES  AND    ITS    DERIVATIVES,  THE   TRIANGLE    OF   FORCES, 
THE   MAGNITUDE    POLYGON,    AND   THE   STRING    POLYGON. 

29.  Demonstration  of  the  Parallelogram  of  Forces 22 

30.  Triangle  of  Forces 24 

31.  Polygon  of  Forces.     Location  of  Resultant  of  Inclined  Forces.     Magnitude 

Polygon 25 

32.  String  Polygon 26 

33.  Additional  Remarks  on  the  String  Polygon 30 

CHAPTER  IV. 

ALGEBRAIC     AND     GRAPHIC     STATEMENTS   OF   THE   CONDITIONS     OF     EQUILIBRIUM 
WITH   APPLICATIONS. 

34.  Nature  of  Statical  Problems 31 

35.  Condition  of  Equilibrium 31 

36.  Algebraic  Statement   of  (A)  and  (2?)  (the  Conditions  of  Equilibrium)  for 

Coplanar  Forces 32 

37.  Graphic  Interpretation  of  (A)  and  (B) 34 

38.  Exercises  in  the  Composition  of  Forces 35 

Exercises  3-7 38 

39.  Generalization  of  the  Three  Classes  of  Resultants 39 

40.  Establishment  of  Equilibrium  by  Use  of  Moments  Alone 39 

41.  Determination  of  Magnitudes  by  Single  Moment  Equations  Alone 41 

42.  Convention  as  to  Algebraic  Signs  in  Moment  Equations 43 

43.  Six  Methods  of  Stating  the  Conditions  of  Equilibrium 43 

CHAPTER  V. 

SCOPE   OF   PURE   STATICS. 

44.  General  Survey  of  the  Scope  of  Pure  Statics 45 

45.  The  Four  Cases 47 

CHAPTER  VI 

SOLUTIONS    OF    STATICAL    PROBLEMS,    WITH   SPECIAL    REFERENCE   TO    THE    FOUR 
MOST    IMPORTANT    CASES. 

46.  The  Solution  of  Statical  Problems 49 


CONTENTS.  vii 

• 

SECTION  PAGE 

47.  Solution  of  Case  1 50 

48.  Solution  of  Case  2a 50 

48  a.  Solution  of  Case  2b 51 

49.  Solution  of  Case  3 53 

50.  Solution  of  Case  4 55 

50  a.  Remarks  on  Cases  3  and  4 58 

Exercises  8-15 58 

CHAPTER  VII. 

ADDITIONAL   GENERAL  TOPICS   AND   PROCESSES. 

5 1.  Graphic  Representation  of  the  Moment  of  a  Force 62 

52.  String  Polygon  for  Parallel  Forces  a  Diagram  of  Moments 63 

53.  Remarks  on  the  String  Polygon  as  a  Diagram  of  Moments 65 

54.  To  Pass  a  String  Polygon  Through  Three  Given  Points 66 

Exercise  16 70 

55.  An  Alternative  View  of  the  Rays  and  the  String  Polygon 70 

PART   II. 

APPLICATIONS. 
CHAPTER  VIII. 

CENTERS   OF  GRAVITY. 

56.  Center  of  Gravity 72 

Exercise  17 77 

CHAPTER  IX. 

STRESS. 

57.  External  and  Internal  Forces 78 

58.  Stress 78 

59.  Kinds  of  Stress 79 

60.  Combined  Stresses 79 

61.  Further  Particulars  Relating  to  Stress 80 

CHAPTER  X. 

ST  RU  CT  U  RE  S  . 

62.^  Definitions 83 

63.  Extent  of  Approximation  to  True  Frames  in  Practice , 84 

64.  Loads  Applied  Elsewhere  than  at  Joints 85 

65.  Frames  in  General 86 

66.  Loads 88 

67.  Stresses  in  Structures 88 

CHAPTER  XL 

STRESSES   IN   NON-FRAMED  STRUCTURES. 

68.  Stresses  in  Non-Framed  Structures 90 

Exercise  18 91 


vin  CONTENTS. 

SECTION  PAGE 

69.  Shear  Diagrams 91 

70.  Flexure  Diagrams 92 

Exercise  IQ 93 

Jri.  Connection  Between  Shear  and  Change  in  Flexure 93 

Exercise  20 95 

CHAPTER  XII. 

STRESSES   IN   FRAMED  STRUCTURES. 

72.  Stresses  in  Framed  Structures 96 

73.  Method  of  Sections 97 

74.  Method  for    Determining  all    the    Stresses    in  a  Frame  under  a  Given 

Load 99 

75.  Example .  100 

76.  Stress  Diagrams 103 

77.  General  Instructions  Regarding  Exercises  Involving  Stress  Diagrams. . . .  104 

78.  Special  Instructions  Regarding  Exercises  21-23 104 

Exercises  21-23 104 

CHAPTER  XIII. 

ADDITIONAL  TOPICS   AND  EXAMPLES. 

79.  Complications  in  Connection  with  the  Analysis  of  Frames 106 

80.  Reactions  due  to  Non- vertical  Forces 107 

81.  The  Fink  Truss 108 

Exercise  24 108 

82.  Triangular  Frame  with  Trussed  Top  Chord 108 

Exercis.6  25 109 

83.  Counters 109 

Exercise  26 no 

84.  Bent  of  a  Mill  Building no 

Exercise  27 ; . , 114 

85.  Cantilever  Bridge 115 

Exercise  28 1 15 

86.  Three-Hinged  Arch 116 

87.  Line  of  Pressure 1 18 

Exercises  29,  30 119 

88.  Hammer  Beam  Truss 120 

Exercise  31 , 122 

89.  Stresses  Due  to  Moving  Loads 123 

90.  Stability  of  a  Masonry  Dam 124 

Exercise  32 125 

91.  Action  and  Reaction  Not  Necessarily  Normal  to  the  Surface  of  Contact. . .  125 

92.  Friction 126 

APPENDIX 130 


STATICS. 


PART  I. 
GENERAL  PRINCIPLES  AND  METHODS. 


INTRODUCTION. 

i.  Algebraic  and  Graphic  Methods  Compared. — There 
are  two  methods  of  representing  quantities  for  the  purpose  of 
computation,  one  in  which  numerals  or  letters  are  used — the 
algebraic  or  so-called  analytical  method — and  the  other  in 
which  quantities  are  represented  by  the  lengths  and  relative 
positions  of  lines — the  graphic  method. 

The  quantities  of  higher  mathematics,  such  as  imaginaries 
and  infinitesimals,  can  be  treated  algebraically  only.  On  the 
other  hand,  many  complicated  relations,  such  as  profiles  of  rail- 
road lines  and  the  varying  pressure  in  a  steam  cylinder,  can 
be  comprehended  and  made  subjects  of  computation  or  measure- 
ment only  by  graphic  representation. 

The  great  bulk  of  ordinary  arithmetic  and  trigonometric 
calculation  can  be  done  by  either  method,  and  which  of  the  two 
should  be  selected  is  purely  a  matter  of  expediency.  In  the 
fundamental  operations  of  arithmetic — addition,  subtraction, 
multiplication,  division,  involution,  and  evolution — the  graphic 
method  is  either  so  simple  and  obvious  that  it  is  rarely  thought 


2  STATICS. 

of  as  a  formal  method  in  connection  with  them,  or  else  so  dis- 
advantageous as  to  be  little  used. 

Problems  involving  much  finding  and  summing  of  products 
or  involving  trigonometric  or  complicated  geometric  relations 
are  found  to  be  an  advantageous  field  for  the  use  of  graphic 
methods,  and  in  this  field,  graphic  methods  assume  a  place  of 
relative  importance. 

Problems  in  equilibrium  are  of  precisely  this  sort,  and 
graphic  methods  of  dealing  with  them  have  been  developed  very 
extensively,  and  even  named,  as  a  body,  Graphical  Statics,  in 
distinction  from  the  body  of  algebraic  methods  in  statical  prob- 
lems— called  Analytical  or  Algebraic  Statics.  All  these 
terms,  as  pointed  out  in  the  preface,  seem  objectionable  to  the 
writer  and  will  not  be  used  further  in  this  work.  The  terms 
algebraic  method  and  graphic  method  will  be  used  in  their 
stead. 

The  underlying  principles  and  the  procedure  in  both  meth- 
ods are,  of  course,  scientifically  correct,  but  they  differ  in  the 
degree  of  precision  attainable.  The  precision  of  results  in 
algebraic  computation  depends  upon  the  extent  to  which  deci- 
mals or  significant  figures  are  carried  out,  and  in  graphical 
work  upon  care  and  skill  in  draughting.  In  the  former  case 
absolute  precision  may  be  approached  as  closely  as  the  com- 
puter may  care  to  go,  but  in  the  latter  early  bounds  are  set  by 
the  limitations  of  draughting  appliances  and  human  eyesight. 
Sometimes  it  is  worth  while  to  use  a  combination  of  the  two 
by  sketching  more  or  less  roughly  the  desired  diagrams  and 
calculating  trigonometrically  instead  of  scaling  the  lengths  of 
resulting  lines.  This  of  course  is  the  algebraic  method  itself 
so  far  as  precision  goes. 

Whatever  may  be  said  about  their  relative  precision  and 
scope,  it  is  certainly  true  that  neither  method  can  yield  a  result 
more  accurate  than  the  data  from  which  it  is  derived.  In 


INTRODUCTION.  3 

A 

structural  design,  the  data  are  rarely  or  never  known  to  a  de- 
gree of  accuracy  greater  than  the  degree  of  precision  readily 
obtained  in  graphic  computation,  so  that  actually,  from  the 
point  of  view  of  the  designing  engineer,  one  method  may 
be  said  to  yield  results  as  trustworthy  as  the  other.  As 
already  stated,  the  two  methods  are  of  equal  correctness  in 
principle. 

The  choice  between  one  method  or  the  other  for  a  given 
problem  before  an  engineer  is  a  matter  of  convenience,  but  in 
many  cases  both  should  be  used  for  the  sake  of  the  check  upon 
results  thus  obtained. 

The  graphic  method  has  the  advantage,  sometimes  impor- 
tant, that  it  presupposes  a  knowledge  of  little  or  no  mathe- 
matics beyond  the  elements  of  plane  geometry,  while  the 
draughting  appliances  required  in  its  use  are  of  the  simplest. 

2.  History  of  Graphic  Methods  in  Statics. — Knowl- 
edge of  the  graphic  methods  in  statics  and  of  their  importance 
and  scope  is  due  mainly  to  the  efforts  of  the  late  Professor 
Culmann  of  Zurich.  Prior  to  1866,  when  he  published  his  great 
work,  "Die  Graphische  Statik,"  the  subject  had  attracted 
little  attention  and  there  existed  only  scattered  and  frag- 
mentary writings  on  it.  Culmann  died  after  completing  the 
first,  only,  of  his  contemplated  volumes,  but  he  had  thor- 
oughly established  the  subject,  and  interest  in  the  graphic 
treatment  of  statical  problems  steadily  spread  through  the 
engineering  world.  Among  other  names  to  be  prominently 
associated  with  the  development  of  the  graphic  methods  are 
Bow,  Maxwell,  Mohr,  W.  Ritter,  Cremona,  Miiller-Breslau, 
Levy. 


CHAPTER   I. 
DEFINITIONS  AND  PRELIMINARIES. 

3.  Mechanics  is  the  science  which  treats  of  rest  and 
motion.  It  is  subdivided  into  Kinematics  and  Dynamics. 

Kinematics  treats  of  motion  apart  from  its  causes.  Its 
scope  is  accordingly  limited  to  the  consideration  of  the  paths 
of  moving  bodies  and  of  that  aspect  of  velocity  and  accelera- 
tion which  involves  only  the  relations  between  space  and  time. 

Dynamics  treats  of  the  effect  of  forces  upon  rest  or  motion. 
It  is  subdivided  into  Kinetics  and  Statics.  Kinetics  deals 
with  the  cases  in  which  given  forces  produce  a  change  in  a 
body  with  respect  to  rest  or  motion,  and  Statics  with  the  cases 
in  which  no  such  change  is  produced. 

Kinetics  accordingly  deals  with  the  velocity  and  accelera- 
tion of  given  bodies  resulting  from  given  forces,  and  with  such 
topics  as  work  and  energy,  momentum,  centrifugal  force, 
impact,  etc.  Statics  is  concerned  exclusively  with  the  con- 
ditions under  which  a  body  under  the  action  of  forces  will 
remain  at  rest  or  undergo  no  change  of  motion.* 

Statics — henceforth  the  exclusive  subject  of  this  book — is 
accordingly  the  science  of  equilibrium,  of  stability — the  science 
by  the  aid  of  which  are  determined  the  forces  necessary  to 

*  The  four  terms  Kinematics,  Dynamics,  Statics,  and  Kinetics  have  an  etymo- 
logical fitness  which  is  worth  noticing.  Kinematics  is  from  Gr.  kinema,  motion; 
Dynamics  from  Gr.  dynamis,  force;  Kinetics  from  Gr.  kinctikos,  putting  in  motion; 
and  Statics  from  Gr.  statikos,  causing  to  stand. 

4- 


DEFINITIONS  AND  PRELIMINARIES.  5 

0 

maintain  a  body  undisturbed  in  its  rest  or  motion  in  spite  of 
disturbing  tendencies,  the  forces  which  must  interact  between 
the  various  parts  of  the  body  to  prevent  its  disruption  under 
the  action  of  given  forces,  and  moreover  the  position  in  which 
a  body  or  system  of  bodies  under  a  given  set  of  forces  will  be 
at  rest. 

4.  Rigid  Body.  Particle. — A  ri^id  body  is  a  body  con- 
ceived to  be  incapable  of  change  in  shape  or  size  under  the 
action  of  forces.  Such  a  body  could  be  affected  only  as  a 
whole  by  forces  which  act  upon  it.  Hence  to  state  that  the 
body  concerned  in  any  problem  is  rigid  or  is  to  be  so  consid- 
ered is  equivalent  to  stating  that  no  thought  is  expected  to  be 
given  to  the  possibility  of  the  problem  being  complicated  by 
the  deformation  or  rupture  of  the  body. 

Although  no  such  thing  as  a  rigid  body  exists  in  nature, 
all  solids  approximate  rigidity  to  a  greater  or  less  extent.  In 
the  case  of  solids  subject  only  to  forces  well  within  their  capac- 
ity to  withstand  them,  such  as  all  properly  designed  engineer- 
ing structures,  the  approximation  to  rigidity  is  very  close,  and 
for  the  purposes  of  statics  no  material  error  results  from  as- 
suming complete  rigidity  in  such  cases. 

A  complete  examination  into  the  safety  of  an  engineering 
structure  involves  (a)  the  determination  of  the  forces  acting  on 
the  body  or  transmitted  through  its  various  parts ;  and  (&)  the 
study  as  to  whether  the  body  and  its  parts  are  able  to  resist 
such  forces  without  rupture  or  undue  deformation.  In  the  first 
step  the  body  is  assumed  rigid,  because  the  results  of  the  step 
are  practically  the  same  as  if  the  body  were  rigid,  and  in  the 
second  the  lack  of  rigidity  is  clearly  recognized  and  provided 
against.  The  first  step  falls  within  the  domain  of  statics,  and 
the  second  within  that  of  the  kindred  science,  Resistance  of 
Materials. 

In  statics  it  is  the  general  practice  to  treat  bodies  as  if  they 


6  STATICS. 

were  rigid,  unless  distinctly  stated  to  the  contrary, — actual 
lack  of  rigidity  being  taken  into  account  only  as  an  aid  in  the 
solution  of  certain  complex  problems  which  .would  otherwise 
be  indeterminate. 

A  particle  is  a  rigid  body  of  the  smallest  conceivable  di- 
mensions. It  is  also  called  a  material  point,  or,  for  short, 
simply  a  point. 

Every  problem  involving  forces  implies  the  existence  of 
bodies  or  particles  on  which  the  forces  act. 

5.  Rest  and  Motion. — Rest  is  the  relation  existing  be- 
tween two  points  when  a  line  connecting  them  does  not  change 
either  in  length  or  in  direction.  If,  on  the  other  hand,  the  line 
does  change  in  either  length  or  direction,  the  relation  between 
the  points  is  motion.  This  line  can  change  in  one  or  both  of 
these  two  particulars  only,  and  motion  can  accordingly  be 
divided  into  two  corresponding  classes  only,  translation  and 
rotation. 

If  the  line  between  two  points  changes  in  length,  one  point 
is  in  translation  with  regard  to  the  other.  If  the  change  is  in 
direction,  one  point  is  in  rotation  about  the  other. 

Two  bodies  are  at  rest  with  regard  to  each  other  when 
every  point  of  one  is  at  rest  with  regard  to  every  point  of  the 
other. 

Two  bodies  are  in  motion  with  regard  to  each  other  when 
any  point  of  one  is  in  motion  with  regard  to  any  point  of  the 
other. 

The  motion  of  a  body  is  translation  when  any  point  of  that 
body  describes  a  straight  line.  If  the  points  of  the  body  de- 
scribe a  series  of  straight  lines,  the  motion  is  pure  translation. 

The  motion  of  a  body  is  rotation  when  such  points  as 
change  in  position  at  all  describe  sets  of  concentric  circles  in 
parallel  planes. 

The  common  normal  to  these  planes  which  contains  the 


DEFINITIONS  AND  PRELIMINARIES.  7 

0 

centers  of  all  the  circles  is  called  the  axis  of  rotation.  This 
axis  may  or  may  not  traverse  the  body,  but  any  points  not 
changing  in  position  must  be  contained  in  it.  If  the  axis  of 
rotation  is  fixed  in  position,  the  motion  is  pure  rotation. 

Of  course  there  can  be  compound  translation,  as  that  of  a 
body  sliding  across  the  deck  of  a  moving  ship;  or  compound 
rotation,  as  that  of  the  moon  about  its  own  center,  and  also 
about  the  earth;  or  combined  translation  and  rotation,  as  that 
of  a  ball  rolling  in  a  straight  path. 

Two  bodies  at  rest  with  reference  to  each  other  may  be 
moving  with  respect  to  a  third.  Rest,  then,  far  from  meaning 
absolute  motionlessness,  means  only  motionlessness  with 
regard  to  a  definite  body  of  reference,  which  in  our  work  will 
usually  be  understood  to  be  the  earth. 

It  may  be  added  that  translation  can  be  regarded  as  the 
extreme  case  of  rotation  in  which  the  axis  is  at  an  infinite 
distance,  and  instead  of  two  subdivisions  of  motion,  rotation 
alone  might  be  regarded  as  covering  the  whole  ground.  This 
point  of  view,  however,  will  not  be  convenient  for  the  purpose 
now  in  hand. 

6.  Force, — Force  is  an    interaction  between   two   bodies, 
either  causing  or  tending  to  cause  a  change  in  their  relative 
rest   or    motion.      A  force  may  always   be  conceived  of  as  a 
push  or  a  pull. 

Forces  are  designated  by  single  capital  letters  as  P,  Q,  R, 
etc.  In  graphical  work  more  elaborate  designations  are 
needed,  which  will  be  explained  in  Chapter  II. 

7.  Elements  of  a  Force. — In  order  that  a  force  may  be  fully 
known,  three  characteristics  of  it,  which  will  be  spoken  of  as 
elements,  are  essential  and  sufficient,  viz. : 

1.  Magnitude. 

2.  Direction  (including  sense). 

3.  Point  of  Application. 


8  STATICS. 

8.  Magnitude. — The  magnitude  of  a  force  is  given  by  stat- 
ing numerically  the  ratio  of  its  effectiveness  in  producing  trans- 
lation  to   that    of   some   arbitrarily   selected    standard    force, 
usually  the  f6rce  of  gravitation  on  a  certain  mass  of  standard 
material  at  a  certain  level  and  latitude;  i.e.,  to  the  amount 
of  force  exerted  by  gravitation  on  the  standard  pound  or  other 
standard  weight.     In  other  words,  the  magnitude  of  a  force 
is  stated  in  units  of  weight. 

The  absolute  unit  of  force,  that  is  the  force  which,  acting 
for  a  unit  length  of  time  on  a  unit  mass  of  material,  will  give 
rise  to  a  unit  velocity,  is  not  used  in  statics. 

9.  Direction  and  Sense. — The  direction  of  a  force  is  the 
direction   of  the   line   along  which  the  force   tends  to  produce 
motion,  i.e.,  of  the  line  of  action  of  the  force.      Direction  in- 
cludes the  sense  of  the  force  as  well  as  the  slope  of  its  line  of 
action.      By  sense  is  to  be  understood  the  specification  as  to 
which    of  the  two  ways   along  the  line    of   action    the    force 
tends. 

Sense  is  the  distinction  between  forward  or  back,  up  or 
down,  ,to  right  or  left  on  a  given  line.  It  is  merely  a  matter 
of  algebraic  sign,  requiring  no  separate  equation  for  its  deter- 
mination, and  hence  is  not  ranked  as  an  element. 

10.  Point  of  Application. — The  point  of  application  is  the 
place  (treated  as  a  point)  upon  a  body  where  the  force  is  brought 
to  bear.      This  point  is,  of  course,  upon  the  line  of  action  and, 
together  with  the  slope,  it  locates  that  line. 

In  the  case  of  rigid  bodies,  any  point  whatever  on  the  line 
of  action  of  the  force  may  be  taken  as  the  point  of  application.* 

Change  in  the  rJoint  of  application  of  a  given  force  affects 
the  body  only  as  to  rotation,  not  at  all  as  to  translation. 

*  This  will  be  seen  upon  adding  anywhere  in  the  line  of  action  of  the  force  two 
opposite  forces  coincident  with  it  and  both  equal  to  it  in  magnitude.  The  three, 
obviously  equivalent  to  the  original,  amount  simply  to  a  new  one  just  like  it. 


DEFINITIONS  AND  PRELIMINARIES.  9 

• 

11.  Coplanar   and     Non-Coplanar    Forces Forces    are 

spoken  of  as   coplanar  and   non-coplanar  according  as  their 
lines  of  action  do  or  do  not  lie  in  one  plane. 

Except  where  distinctly  stated  to  the  contrary,  and  except 
where  statements  are  evidently  general  (as  in  the  whole  of  this 
chapter),  coplanar  forces  will  form  the  exclusive  subject  or 
these  pages.  This  limitation  conduces  to  simplicity  without 
real  loss  of  generality,  for  coplanar  forces  once  understood, 
the  treatment  of  non-coplanar  forces  will  offer  little  difficulty 
to  the  unaided  reader. 

12.  Concurrent  and  Non- Concurrent  Forces. — Forces  are 
spoken  of  as  concurrent  or  non-concurrent  according  as  their 
lines  of  action  do  or  do  not  meet  in  a  point.      Of  course,  either 
may  or  may  not  be  coplanar.      The  study  of  concurrent  forces 
is  sometimes  spoken  of  as  statics  of  a  particle,  and  of  non- 
concurrent  forces  as  statics  of  a  rigid  body. 

Whether  a  set  of  forces  is  concurrent  or  not  is  a  matter 
affecting  rotation  only,  the  translatory  effect  of  a  force  being 
independent  of  its  point  of  application. 

13.  Equilibrium. — Equilibrium  is  that  condition  of  a  set 
of  two  or  more  forces  the  result  of  which  is  that  their  combined 
effect  on  a  body  produces  no  change  in  the  body  with  respect 
to  rest  or  motion. 

A  body  is  said  to  be  in  equilibrium  when  it  is  acted  on  by 
a  set  of  forces  in  equilibrium.  A  body  in  equilibrium  may  be 
either  at  rest  or  moving  with  uniform  speed  (either  of  transla- 
tion or  rotation)  with  regard  to  the  body  of  reference.  If  it 
be  moving  in  either  of  the  ways  just  mentioned,  it  is  the  result 
of  the  previous  application  of  an  unbalanced  force  or  of  un- 
balanced forces,  and  any  forces  actually  affecting  it  while  in 
such  motion  are  forces  under  the  action  of  which  a  body  could 
equally  well  be  at  rest. 

A  set  of  forces,  therefore,  under  which  a  body  will  be  at 


io  STATICS. 

rest  is  always  a  set  in  equilibrium;  or,  in  other  words,  condi- 
tions  of  rest  are  always  conditions  of  equilibrium.  It  follows, 
that,  by  establishing  the  conditions  of  rest,  the  familiar  most 
usual  phase  of  equilibrium,  we  establish  the  general,  universal 
conditions  of  equilibrium. 

14.  Equivalence. — Two    forces  or  sets    of  forces   having 
identical  effects  on  the  rest  or  motion  of  a  body  both  with  re- 
spect to  translation  and  to  rotation  are  termed  equivalent. 

15.  Resultant  and  Equilibrant.     Composition  of  Forces. 
— A  single  force  equivalent  to  a  set  of  forces  is   called   the 
resultant  of  the  set.      A  single  force  the  addition  of  which 
to  a  set  of  forces  produces  equilibrium  is  called  the  equilibrant 
of  that  set  of  forces.      The  resultant  and  the  equilibrant  for  a 
given  set  differ  in  sense  only,  i.e.,  one  is  simply  the  equal  and 
opposite  of  the  other.      Of  a  set  of  forces  in  equilibrium  any 
one  of  them  is  the  equilibrant  of  all  the  rest.      The  operation 
of  finding  the  resultant  or  equilibrant  of  a  set  of  forces  is  called 
Composition  of  Forces. 

16.  Components. — Any  one  of  a  set  of  forces  having  a 
given  resultant  is  a  component  of  that  resultant.      Evidently 
there  may  be  any  number  of  components  of  a  force. 

Rectangular  components  of  a  force  are  two  components 
equivalent  to  it  whose  lines  of  action  are  at  right  angles  to 
each  other. 

A  component  in  a  given  direction  is  understood  to  be  the 
one  of  a  pair  of  rectangular  components  whose  line  of  action  is 
parallel  to  the  given  direction.  Such  a  component  measures 
the  total  tendency  of  the  given  force  to  produce  motion  in 
the  given  direction. 

17.  Relation  between  Two  Components  and  their  Result- 
ant.    Parallelogram  of  Forces.* — If  the  adjacent  sides  of  a 
parallelogram  represent  two  concurrent  forces  in  magnitude  and 
*  See  also  §  29  for  proof  and  additional  discussion  of  the  parallelogram  of  forces. 


DEFINITIONS  AND  PRELIMINARIES.  II 

direction,  the  diagonal  of  this  parallelogram  starting  at  the  in- 
tersection of  these  two  sides  will  represent  their  resultant  in 
magnitude  and  direction,  provided  the  two  forces  act  either 
both  toward  or  both  away  from  this  intersection.  The  result- 
ant will  then  also  act  toward  or  away  from  the  point  of  inter- 
section as  the  case  may  be.  Ordinarily  the  components  are 
regarded  as  acting  away  from  their  common  point,  and  accord- 
ingly the  resultant  also. 

Corollary. — If  a  be  the  inclination  of  any  force,  P,  to  any 
given  line,  the  components  along  and  at  right  angles  to  this  line 
will  have  the  magnitudes  P  cos  a  and  P  sin  a  respectively. 

EXERCISE  i.*  COMPONENTS.  Find  the  components,  both  algebra- 
ically and  graphically,  of  a  force  P  whose  magnitude,  direction,  and 
point  of  application  are  20  Ibs.,  210°,  and  (2,  6)  respectively,  along  the 
axes  of  X  and  Y,  and  also  along  five  lines,  Mi  Ni,  M*  N*t  M»  Ns,  M*  N*, 
M*  A'»,  inclined  to  the  axis  of  Xby  45°,  60°,  120°,  45°,  and  30°  respec- 
tively, and  which  pass  through  points  (o,  o),  (-6,  o),  (6,  o),  (8,  o),  and 
10,  o)  respectively. 

SUGGESTIONS  AS  TO  GRAPHIC  WORK. 

Use  as  scales  i  in.  =  10  Ibs.  and  4  units  of  length.  Plot  the  line  of 
action  of  P  in  its  proper  relation  to  the  preferably  horizontal  and  vertical 
axes  of  coordinates  and  the  given  lines  Mi  N\  .  .  .  M*,  Ni.  This  will  serve 
as  a  diagram  of  data  for  both  methods.  The  problem  can  now  well 
be  stated  in  condensed  form  across  the  top  of  the  sheet. 

Show  P  apart  from  this  diagram  with  the  proper  direction  and  with 
the  magnitude  marked  off  on  it  to  scale.  At  a  little  distance  from  it, 
draw  parallels  to  all  the  given  lines.  Construct  the  required  com- 
ponents, scale,  and  show  results  in  their  proper  places  with  dimension 

lines. 

SUGGESTIONS  AS  TO  ALGEBRAIC  WORK. 

Determine  the  angles  between  the  line  of  action  of  P  and  each  of  the 
seven  given  lines.  Proceed  with  the  necessary  computations  using  this 
form: 

0.707 
For  Mi  Ni,     20  cos  45°  =  14.14. 

"    M*  N*t  etc. 

*  For  the  solution  of  this  and  all  the  following  exercises,  sheets  of  paper  about 
I2X1^  inches,  preferably  ruled  in  inch  and  tenth-inch  squares,  are  recommended 
as  affording  sufficient  room,  and  as  convenient  in  shape  for  the  double  solutions 
usually  required. 


12  STATICS. 

Finally  collect  results  and  show  them  in  parallel  columns  thus : 

Line  Gr.  Al. 

Mi  Ni  14.1  14.14 

QUESTIONS  ON  EXERCISE  i. 

How  does  the  line  MiNi  differ  from  MtN4  ? 

What  difference  does  this  cause  in  the  components  along  these  two 
lines? 

How  would  changing  the  point  of  application  of  P  to  say  —3,  5,  or 
elsewhere,  affect  the  results  of  this  exercise? 

From  the  seven  components  found,  can  two  be  selected  which  would 
fully  replace  P  if  properly  located  ?  Is  more  than  one  such  selection 
possible  ? 

What  would  have  to  be  the  location  of  such  components  in  order  that 
they  might  replace  P  ? 

18.  Couples  and  their  Moments. — Two  equal  and  oppo- 
site forces  with  parallel  lines  of  action  constitute  a  couple. 
The  sole  tendency  of  a  couple  is  to  produce  rotation.  The 
measure  of  a  couple,  which  can  be  nothing  else  than  the 
measure  of  its  tendency  to  produce  rotation,  is  called  the 
moment  of  the  couple. 

It  has  been  observed  as  one  of  the  fundamental  laws  of 
matter  that  the  moment  of  a  couple  is  proportional  alike  to  the 
common  magnitude  of  the  forces  of  the  couple  and  to  the 
shortest  distance  between  their  lines  of  action.  This  distance 
is  called  the  arm  of  the  couple.  The  product  of  such  magni- 
tude and  arm  expresses  the  moment  of  the  couple. 

The  unit  couple  consists  of  two  forces  of  unit  magnitude 
with  unit  arm,  and  its  moment  is  the  unit  for  measuring 
couples.  This  unit  is  called,  in  the  English  system,  inch- 
pound,  foot-pound,  foot-ton,  etc.  It  must  not  be  confounded 
with  the  units  of  work  and  energy  with  the  same  names. 

The  sense  of  a  couple  will  be  called  positive  or  negative 
according  as  its  tendency  is  clockwise  or  not. 

The  plane  of  a  couple  once  known  nothing  further  need  be 


DEFINITIONS  AND  PRELIMINARIES.  13 


known  of  it  but  its  moment,  including,  of  course,  the  sense  of 
the  possible  rotation. 

Either  of  the  forces  of  a  couple  can  be  treated  for  computa- 
tion exactly  like  one  of  any  other  system  of  forces. 

The  effect  of  a  couple  is  the  same  wherever  in  its  plane  or 
in  a  parallel  plane  its  forces  may  be  applied. 

Couples  in  the  same  plane  can  be  summed  by  summing 
their  moments,  and  a  couple  equivalent  to  the  set  can  be  thus 
evaluated,  just  as  forces  in  the  same  straight  line  can  be 
summed  by  summing  their  magnitudes  with  the  effect  of 
evaluating  a  force  equivalent  to  the  set.  In  such  work  sense 
must,  of  course,  be  duly  regarded. 

19.  Center  of  Rotation  for  a  Couple. — The  center  of  rota- 
tion for  a  couple  may  be  anywhere  in  its  plane,  and  is  deter- 
mined only  by  the  relative  motions  of  the  bodies  from  which 
come  the  forces  constituting  the  couple. 

An  interesting  and  important  special  case  is  that  in  which 
only  one  of  the  forces  can  be  maintained  in  action  at  successive 
points  of  application.  The  center  of  rotation  will  then  be  the 
point  of  application  of  the  other  force,  a  familiar  illustration 
being  a  pulley  and  belt.  The  point  of  application  of  the  resist- 
ance of  the  shaft  cannot  change,  while  that  of  the  force  from 
the  belt  retreats  freely  as  the  pulley  turns.  The  center  of 
rotation  is  accordingly  at  the  center  of  the  shaft. 

20.  Single  Force  Always  Replaceable  by  a  Single  Force 
and. a  Couple. — Let  P  (Fig.  i)  be  any  force  applied  to  a  body 
and    m    be    any  point    in    the    body. 

Suppose  two  additional  forces,  P'  and 

P'1 ',   opposite    in   sense,   parallel    to    P 

and  of  the  same  magnitude    as  P,   to 

be  applied  to  the  body  at  m,  distant  p 

from  the  line  of  action  of  P.     Evidently  FlG- 

the  condition  of  the  body  is  unchanged,  but  the  three  forces 


14  STATICS. 

are  seen  to  be  equivalent  to  P  with  a  new  point  of  application 
m,  and  a  couple  of  value  P  X  P- 

Hence  it  may  be  stated  that  any  force  P  can  be  replaced 
by  ax  force  of  the  same  magnitude  and  direction  distant/  from 
its  original  position,  and  a  couple  of  the  value  P  x  P>  In 
other  words,  points  of  application  of  forces  can  be  shifted  at 
will  without  effect  on  equilibrium,  if,  at  each  change,  a  suit- 
able couple  be  added  to  the  system. 

Any  set  of  non-concurrent  forces  can  thus  be  reduced  to  an 
equivalent  set  of  couples  and  concurrent  forces. 

21.  Moment  of  a  Force. — The  moment  of  a  force  with  re- 
spect to  a  point  is  the  name  given  the  moment  of  the  couple 
consisting  of  the  given  force,  and  a  force  equal,  opposite,  and 
parallel  to  it  conceived  to  be  acting  at  the  given  point. * 

The  arm  of  this  couple  is  evidently  the  perpendicular  dis- 
tance from  the  point  to  the  line  of  action  of  the  given  force,  a 
distance  which  may  conveniently  be  called  the  arm  of  the  force 
with  respect  to  the  point. 

The  moment  of  a  force  with  respect  to  a  point  is  accordingly 
expressed  by  the  product  of  the  magnitude  of  the  force  into  its 
arm,  and  is  called  positive  or  negative  according  as  the  con- 
ceived tendency  to  rotation  is  clockwise  or  not. 

The  point  to  which  the  moment  is  referred  is  commonly 
called  the  center  of  moments,  or,  more  briefly,  the  center. 

The  moment  of  a  force  is  sometimes  referred  to  an  axis, 
meaning  thereby  the  axis  of  the  conceived  rotation.  This  axis 
will  be  the  straight  line  through  the  center  normal  to  the  plane 
of  the  force  and  its  arm.  The  moment  with  respect  to  this 

*  Note  that  this  is  the  same  thing  as  saying  that  the  moment  of  a  force  with 
respect  to  a  point  is  the  measure  of  its  tendency  to  produce  rotation  about'the 
point,  assuming  the  point  to  be  fixed.  For  assuming  the  point  to  be  fixed  is 
really  assuming  it  to  be  always  subject  to  a  force  equal,  opposite,  and  parallel 
to  the  given  one.  This  point  of  view  gives  rise  to  the  phrase  moment  of  a 
force  about  a  point. 


DEFINITIONS  AND  PRELIMINARIES.  i$ 

A 

axis  is  also  spoken  of  as  the  moment  with  respect  to  the  plane 
which  contains  this  axis  and  is  parallel  to  the  force. 

EXERCISE  2.  MOMENTS  OF  A  FORCE. — Find  the  moments  in  foot- 
pounds, both  in  magnitude  and  sign,  of  a  force  of  20  Ibs.,  whose  direction 
and  point  of  application  are  respectively  60°  and  (6,  2),  about  the  five 
points  (o,  o),  (—3,  —  6),  (8,  o),  (-  5,  2),  (4,  —  3). 

a.  Obtain  the  results  in  the  most  direct  way,  by  scaling  arms  from  a 
carefully  plotted  diagram  of  data.     Note  that  the  result  necessarily  lacks 
precision  and   is  affected  by  errors  in   draughting.     Scale  i  in.  =  2  ft. 
Show  scaled  dimensions  in  the  diagram. 

b.  Obtain  the  results  by  adding  algebraically  obtained  moments  of 
components  of  the  given  force — a  purely  algebraic  process.     Note  that 
the  results  are  wholly  independent  of  draughtsmanship  and  make  a  rigid 
check  of  results  of  (a)  and  can  be  made  as  precise  as  the  nature  of  the 
case  permits. 

Do  the  work  in  this  form  : 

/,  =  20  x  ?  =  ? 

/a  =  20    X  ?  =  ? 

etc. 

0.500  0.866 

(  Mi  =  20  cos  60°  x  2  —  20  sin  60°  x  6  =  20.00  —  103.92  =  —  83.92 
&<  10.0  5-196 

( Mi  =       etc. 

Record  results  in  this  form  : 


Ml 


22.  Propositions  Regarding  the  Moment  of  a  Force.  — 

There  follow  at  once  from  the  definition  of  the  moment  of  a 
force  the  three  following  important  propositions. 

I.  For  a  force  greater  than  zero,  the  moment  can  be  zero 
only  with  respect  to  a  point  in  its  own  line  of  action. 

II.  A  zero  moment  with  respect  to  a  point  off  a  stated  line 
of  action  can  result  only  from  a  zero  force  in  that  line. 

III.  A  zero  force  with  an  infinite  arm  can  have  a  finite 
moment. 


1 6  STATICS. 

The  "-zero  force  with  an  infinite  arm  "  of  the  last  proposi- 
tion is  simply  a  way  of  expressing  any  couple,  as  will  be  seen 
later. 

23.  Proposition. — The  sum  of  the  moments  of  two  equal, 
opposite,  and  parallel  forces  (i.e.,  the  moment  of  a  couple)  is 
constant  for  all  points  in  their  plane. 

This  the  reader  can  readily  prove  for  himself  by  taking  any 
center  and  summing  for  it  the  moments  of  the  two  forces,  using 
any  convenient  letters  for  their  arms. 

24.  Sets  of  Forces  Classified. — It  follows  from  §  5  that 
all  possible  conditions  of  a  body  with  respect  to  rest  and  motion 
fall  within  one  of  four  classes,  viz. : 

1.  Translation. 

2.  Rotation. 

3.  Both  Translation  and  Rotation. 

4.  Neither  Translation  nor  Rotation. 

Any  set  of  forces  must  accordingly  be  such  as  will  produce 
one  of  these  four  conditions.  We  know  by  observation  that 
a  single  force  will  produce  translation  and  that  a  couple  will 
produce  rotation.  Knowing  also  that  translation  and  rotation 
can  also  result  from  sets  of  forces,  we  unhesitatingly  conclude 
that  for  producing  translation,  a  set  could  be  replaced  by  some 
single  force  and  for  producing  rotation,  by  some  couple. 
Translation  being  the  result  of  a  single  force  or  a  set  equiva- 
lent thereto,  and  rotation  the  result  of  a  couple  or  a  set 
equivalent  thereto,  it  follows  that  any  set  of  forces  must  fall 
into  one  of  four  classes  corresponding  to  one  of  the  four  con- 
ditions of  a  body  mentioned  at  the  outset  of  this  section. 

Any  set  of  forces  is  therefore  reducible  to  one  of  the  follow- 
ing equivalents,  viz. : 

1.  Single  Force, 

2.  Couple, 

3.  Single  Force  and  Couple, 

4.  Neither  Single  Force  nor  Couple, 


DEFINITIONS  AND  PRELIMINARIES.  17 

• 
corresponding  respectively  to  the  four  conditions  of  a  body  just 

mentioned. 

Of  course,  the  last  named  of  these  conditions  is  equilibrium. 

It  should  be  pointed  out,  in  passing,  that  a  force  and  a 
couple  in  the  same  or  parallel  planes  are  reducible  to  a  single 
force,  and  that  the  non-reducible  case  of  a  force  and  a  couple  is 
that  in  which  the  force  is  not  parallel  to  the  plane  of  the  couple. 

25.  Conditions  of  Equivalence. — THEOREM. — If,  for  two 
sets  of  forces  in  a  given  plane,  the  sum  of  the  moments  of  the 
forces  about  three  points  not  in  the  same  straight  line  be  the 
same  for  each  set,  the  two  sets  are  equivalent. 

PROOF. — As  shown  in  the  preceding  section,  each  of  the 
two  sets  must  be  reducible  to  a  single  force,  a  couple,  or  is  in 
equilibrium,  there  being  no  possibility  of  the  non-reducible  case 
of  a  force  and  couple  since  the  forces  are  coplanar.  There  can 
accordingly  be  but  six  different  combinations  of  sets  from  this 
point  of  view.  They  and  their  respective  relations  to  the 
theorem  are  as  follows : 

(a)  Each  set  reducible  to  a  single  force.  — Upon  considera- 
tion of  the  moment  of  a  force,  as  defined  in  §  21,  it  is  apparent 
that  two  forces  can  have  the  same  moment  about  a  point  only 
when  the  point  is  distant  from  their  lines  of  action  inversely  as 
the  magnitudes  of  the  forces.  In  a  given  case  any  point  out- 
side a  definite  straight  line  cannot  meet  these  conditions, 
except  m  the  special  case  when  the  two  forces  are  coincident, 
and  of  the  same  magnitude  and  direction,  i.e.,  when  the  forces 
are  equivalent.  Therefore  equality  of  moments  for  each  of 
three  points  not  in  the  same  straight  line  demonstrates  equiva- 
lence for  this  combination  of  sets. 

($)  Each  set  reducible  to  a  couple. — The  moment  of  a 
couple  (§  23)  being  a  constant  for  all  points  in  its  plane,  equal- 
ity of  moments  for  the  two  sets  for  any  one  point  demonstrates 
equivalence  for  this  combination. 


1 8  STATICS. 

(<:)  Each  set  in  equilibrium. — The  two  sets  are  equivalent 
by  definition,  and  the  sums  of  their  moments  are  bound  to  be 
equal  for  any  and  all  points.  The  constant  value  of  this  sum  is 
of  course  zero. 

(d)  One  set   reducible    to   a   single  force,    the   other  to  a 
couple, — necessarily  non-equivalent. — As   in    combination    I., 
the  moments  here  could  be  equal  only  for  points  on  a  definite 
straight  line.      Their  being  equal  for  each  of  three  points  not  in 
a  straight  line  would  exclude  the  possibility  of  this  combination. 

(e)  One  set  reducible  to  a  simple  force,   the  other  in  equi- 
librium,— necessarily  non-equivalent. — Equality    of  moments 
could  then  arise  only  for  points  on  a  definite  straight  line — the 
line  of  action  of  the  single  force.      Equality  for  each  of  three 
points  not  in  a  straight  line  would  exclude  the  possibility  of  this 
combination. 

(f)  One  set  reducible  to  a  couple,  the  other  to  equilibrium, 
—  necessarily    non-equivalent.  —  Here    equality    of   moments 
could  exist  for  no  point  whatever.      Equality  of  moments  for 
all  points  or  any  point  excludes  the  possibility  of  this  combina- 
tion. 

Since  in  all  of  the  possible  combinations  of  sets  of  forces, 
the  equality  of  the  sums  of  the  moments  of  the  forces  of  the 
two  sets  for  three  points  not  in  the  same  straight  line  is  impos- 
sible without  equivalence  in  the  sets,  the  theorem  is  proved. 

Corollary. — If  the  sets  are  equivalent,  their  moments  are 
equal  for  all  points  in  the  plane  and  conversely. 


CHAPTER    II. 
NOTATION  AND  CONVENTIONS. 

26.  Notation.  Conventions  Regarding  Elements  of 
Forces.  (For  illustrations  see  §  28.) — Magnitude  is  expressed 
in  algebraic  work  by  numerals  or  letters ;  in  graphic  work  by 
lengths  of  lines.  The  capitals  P,  Q,  R,  etc.,  are  used  both 
as  general  designations  of  forces,  and  also  as  the  magnitudes 
of  these  forces  when  the  magnitudes  are  not  given  numerically. 

Magnitudes  may  always  be  regarded  as  essentially  posi- 
tive. If,  in  solving  an  equation,  the  value  of  a  magnitude  ap- 
pears to  be  negative,  it  is  to  be  understood,  nevertheless,  that 
the  force  is  positive  as  usual,  but  that  its  sense  is  opposite  to 
that  assumed  in  stating  the  equation. 

Direction  can  be  expressed  in  algebraic  work  by  the  angle 
(a,  6,  etc.,  or  30°,  60°,  etc.)  which  the  line  of  action  makes 
with,  say,  a  horizontal  line,  and,  by  suitable  convention,  sense 
can  readily  be  included  in  direction.  The  convention  as  to 
sense  which  will  be  used  in  this  work  will  be  the  familiar  one 
of  trigonometry,  in  which  the  angles  are  measured  anti- clock- 
wise from  a  horizontal  axis  continuously  from  o°  to  360°,  the 
measurement  always  being  made  to  that  portion  of  the  line  of 
action  on  which  the  sense  will  be  away  from  the  horizontal 
axis.  Accordingly,  for  example,  a  force  with  direction  30° 
would  be  one  to  the  right  and  upwards;  one  of  210°,  to  the 
left  and  downwards. 

Sometimes  when  the  line  of  action  of  a  force  is  known  but 

19 


20  STATICS. 

the  sense  as  well  as  magnitude  unknown,  it  will,  as  above  in- 
timated, be  convenient,  for  the  algebraic  determination  of  the 
sense,  to  permit  it  to  be  indicated  by  the  sign  of  the  root  of  the 
equation  which  determines  the  magnitude.  The  sense  once 
determined,  it  should  finally  be  associated  with  direction  as 
usual. 

Direction  is  expressed  in  graphic  work  by  the  slope  of  a 
visible  line,  and  sense  is  distinguished  by  an  arrow  placed 
upon  the  line. 

The  point  of  application  is  expressed  in  algebraic  work  by 
coordinates  (one  of  which  it  will  often  be  convenient  to  make 
zero)  and  in  graphic  work  by  points  properly  plotted. 

27.  Space  Diagram  and  Magnitude  Diagram. — To  com- 
pute graphically  with  forces,  two  distinct  but  intimately  related 
diagrams  are  from  the  nature  of  the  case  necessary,  viz.,  the 
Space  Diagram  and  the  Magnitude  Diagram. 

In  the  space  diagram  only  directions  and  points  of  appli- 
cation are  plotted,  and  only  points  of  application  are  determined. 

In  the  magnitude  diagram  only  magnitudes  and  directions 
are  plotted,  and  only  magnitudes  and  directions  determined. 

The  scale  of  the  former  is  that  of  lengths,  as  inches,  cen- 
timeters, etc.,  while  the  scale  of  the  latter  is  the  scale  offeree 
units,  as  pounds,  kilograms,  etc. 

Forces  are  designated  in  the  space  diagram  by  small  let- 
ters on  each  side  of  the  line  of  action,  and  in  the  magnitude 
diagram  by  the  corresponding  capital  letters  at  each  end  of  the 
proper  line.  This  particular  style  of  lettering  is  Bow's  sys- 
te-m.  It  has  great  advantages  over  any  other. 

Other  convenient  characters  may  be  affixed  to  either  dia- 
gram. See  Figs.  2  and  3. 

28.  Illustration   of  Scheme  of  Notation. — The    scheme 
of   notation    enunciated    in    the    two  preceding  sections  may 
be    illustrated    fully    by  taking  two    forces,   Pl    and  Pv  their 


NOTATION  AND   CONVENTIONS. 


21 


magnitudes,  directions,  and  points  of  application  being  as- 
sumed respectively  to  be  20  Ibs.,  240°,  (o,  4),  and  40  Ibs., 
150,  (6,  o). 

Putting  them  in  their  most  condensed  form  algebraically, 
they  may  be  written 

Pl  —  (20  Ibs.  240°  o,  4) 
P2=  (40  Ibs.  1 50°  6,  o). 


Space  Diagram 
Scale,  i  in.  =  8  ft. 

FIG.  2. 


Magnitude  Diagram 
Scale,  i  in.  =  60  Ibs. 

FIG.  3. 


Graphically   the   same  data  would   be  shown  as  in  Figs. 
2  and  3. 


CHAPTER   III. 

PARALLELOGRAM  OF  FORCES  AND  ITS  DERIVATIVES, 
THE  TRIANGLE  OF  FORCES,  THE  MAGNITUDE  POLY- 
GON, AND  THE  STRING  POLYGON. 

29.  Demonstration  of  the  Parallelogram  of  Forces.* 
THEOREM. — If  two  concurrent  forces,  P  and  Q,  both  taken  as 
acting  away  from  their  common  point  M,  be  represented  in 
direction  and  magnitude  by  two  sides  of  a  parallelogram 
meeting  at  M,  their  resultant,  R,  is  represented  in  all  its 
elements  by  the  diagonal  starting  at  M. 

PROOF. — From  the  corollary  of  §  25  it  follows  that  the 
theorem  will  .be  proved  if  it  can  be  shown  that  the  sum  of  the 
moments  of  P  and  Q  about  all  points  O  in  the  plane  is  equal  to 
the  moment  of  R  about  all  such  points  O. 

Let  (Fig.  4)  /,  q,  and  r  be  the  arms  respectively  of  P,  Q, 
and  R  with  reference  to  O  where  O  is  any  point  whatever  in 
their  plane,  and,  noting  the  diversity  in  sign  between  the  mo- 
ments of  P  and  Q,  we  have  only  to  prove  that  Rr  =  Pp  —  Qq. 

*  The  parallelogram  of  forces  was  first  pointed  out  to  the  world  in  1687  by 
Sir  Isaac  Newton  and  Varignon,  probably  independent  of  each  other.  The  fore- 
'going  is  only  one  of  many  known  proofs.  It  is  closely  akin  to  that  of  Rankine, 
Applied  Mechanics,  I5th  ed.,  p.  35.  See  Bowser's  Analytic  Mechanics,  p.  24,  and 
Weisbach's  Mechanik,  I,  p.  165. 

Many  writers  and  teachers  content  themselves  with  the  view  that  the  whole 
proposition  is  a  truth  of  observation  and  experience  quite  as  much  as  the  prem- 
ises which  have  to  be  resorted  to  for  its  proof,  and  so  find  little  or  no  need 
for  mathematical  proofs.  But  when  the  premises  are  materially  more  simple  and 
more  a  part  of  common  experience  than  the  proposition  itself,  the  resulting  proof 
is  believed  to  be  well  worth  the  consideration  of  beginners. 

22 


PARALLELOGRAM  OF  FORCES  AND  ITS  DERIV  AllVhs. 


23 


The  point  common  to  P  and  Q  being  M,  with  coordinates 
x  and  y  referred  to  axes,  OX  and  OY  parallel  respectively  to 


O/  s> 


/ 


FIG.  4. 

P  and  Q,  the  values  of  the  arms  can  be  written  directly  from 
the  figure,  calling  the  inclination  of  P  to  Q  and  R,  GO  and  a 
respectively, 

p  —  y  sin  GJ, 

q  =  x  sin  &>, 

r  =  y  sin  (a?  —  OL)  —  x  sin  <*. 

From  the  figure,  as  a  condition  that  R  may  be  the  diagonal 
of  the  parallelogram,  by  the  theorem  of  sines 


sin  co 


Substituting  these  five  values  in  the  equation  Rr  = 
as  a  means  of  testing  its  validity,  there  results 

Ry  sin  (GO  —  a)  —  Rx  sin  a  = 

R  sin  (GO  —  a]  y  sin  GO       R  sin  a  x  sin  &> 


sin  co 


sin 


24  STATICS. 

or 

y  sin  (GO  —  a)  —  x  sin  a  =  y  sin  (GO  —  a)  —  x  sin  a, 

an  identity,*  and  the  theorem  is  proved. 

Corollary  I.  Three  forces  in  equilibrium  must  meet  in  a 
point. 

Three  forces  may  be  parallel  and  still  be  in  equilibrium — 
a  fact  not  at  variance  with  the  preceding  sentence  if  parallel 
forces  be  considered  a  limiting  case  with  a  common  point  at 
infinity. 

Corollary  II.  With  rectangular  axes,  and  with  (x,  y)  as 
any  point  on  the  line  of  action  of  any  force,  it  appears, 
from  the  value  of  r,  that  the  arm  of  any  force  referred  to  the 
origin  is 

,  .  .-,.—- „,     M»U 

a  =  y  cos  a  —  x  sin  a, 

where  <x  is  the  direction  of  the  force. 

30.  Triangle  of  Forces. — Observe  in  the  parallelogram  of 
forces  of  Fig.  4  that  R  divides  the  parallelogram  into  two  equal 
triangles,  MAB  and  MAC,  either  of  which  shows  fully  the 
relations  between  two  forces  and  their  resultant  as  to  magnitude 
and  direction.  Either  serves  the  same  purpose  as  the  whole 
parallelogram  so  far  as  these  two  elements  are  concerned, 
and  will  be  used  in  place  of  it  under  the  name  triangle  of 
forces. 

Observe  that  if  the  three  forces  are  in  equilibrium  the 
arrows  will  follow  one  another  around  the  triangle,  and  that 
in  case  all  the  arrows  do  not  follow  one  another  the  two  forces 
for  which  they  do  follow  have  the  third  for  a  resultant. 

*  Observe  that,  whatever  the  location  of  M  and  the  direction  of  the  forces  P 
and  Q,  the  signs  connecting  Pp  and  Qq  in  Rr  =  Pp  —  Qq,  and  those  connecting  y 
sin  (GO— a)  and  x  sin  (GO— a)  in  the  expression  for  r  are  always  identical  and  that 
the  identity  here  noted  is  not  dependent  upon  special  conditions. 


PARALLELOGRAM  OF  FORCES  AND  ITS  DERIVATIVES.        25 

w 

Observe  also  in  the  triangle  offerees  that  one  of  the  three 
forces  must  always  be  shown  in  a  line  of  action  not  its  own ; 
in  fact  it  will  usually  be  found  convenient  to  use  parallels  to 
the  actual  lines  of  action  of  all  the  forces.  This  gives  rise  to 
the  magnitude  diagram  of  §  27. 

From  the  triangle  of  forces  it  follows  that  if  a  be  the  incli- 
nation of  a  force  P  to  a  given  line,  P  cos  a  and  P  sin  a 
measure  the  components  of  P  parallel  and  normal  to  that  line 
respectively  (cf.  Cor.,  §  17). 

31.  Polygon  of  Forces,  Location  of  Resultant  of  Inclined 
Forces.  Magnitude  Polygon. — The  resultant  of  any  set  of 
forces  can  be  determined  in  magnitude  and  direction  by  form- 
ing a  polygon  with  its  sides  proportional  to  the  magnitudes  of 
the  given  forces,  and  parallel  to  them,  with  arrows  following 
one  another.  The  magnitude  and  direction  of  the  resultant 
will  be  shown  by  the  closing  line  of  the  polygon,  the  proper 
sense  being  away  from  the  starting-point. 

This  follows  immediately  from  the  triangle  of  forces,  as 
will  be  seen  from  Fig.  5 ,  where  the  process  of  finding  the  mag- 
nitude and  direction  of  the  resultant  of  the  four  forces  ab,  be, 
cd,  and  de  is  shown  in  detail. 

AB  combined  with  BC  by  the  triangle  of  forces,  Fig.  %b, 
leads  to  their  resultant  A  C,  which  is  in  position  to  be  combined 
at  once  with  CD  to  yield  AD — the  desired  resultant  of  AB, 
BC,  and  CD.  This  process  can  obviously  be  continued  indefi- 
nitely. The  partial  resultants  are  seen  to  be  actually  super- 
fluous in  the  construction,  and  the  polygon  ABODE  might  have  ' 
been  built  up  without  them.  This  polygon  is  commonly  called 
the  polygon  of  forces,  but  the  more  distinctive  name,  the 
magnitude  polygon  will  be  preferred  in  this  book. 

Observe  that  while  the  magnitude  polygon  does  not  locate 
the  points  of  application  of  AC,  AD,  or  AE,  it  still  furnishes 
important  aid  in  finding  them.  We  need  only  to  observe  that, 


26 


STATICS. 


in  accordance  with  the  parallelogram  of  forces,  the  point  of 
application  of  A  C  is  at  the  intersection  of  ab  and  be.     Then  ac 


can  be  drawn,  and  where  it  cuts  cd  is  a  point  of  ad.  Drawing 
ad,  where  it  cuts  de  is  a  point  of  aey  and  so  on  for  any  number 
of  forces. 

A  method  applicable  to  parallel  forces  will  be  developed 
presently. 

Corollary.  If  the  last  point  of  the  magnitude  polygon  co- 
incides with  the  first  point,  i.e.  if  the  polygon  closes,  the  result- 
ant is  zero  and  the  set  of  forces  is  equivalent  to  a  couple  or  is 
in  equilibrium,  according  to  the  relative  positions  of  the  given 
forces. 

32.  String  Polygon. — As  has  been  seen,  the  magnitude 
polygon  takes  into  account  only  magnitudes  and  directions  of 
forces.  The  next  step  is  to  establish  the  general  method  by 
which  points  of  application  are  taken  into  account  and  deter- 
mined. This  method  involves  a  construction  in  the  space  dia- 
gram in  close  relation  with  the  magnitude  polygon,  which  may 
be  developed  as  follows. 

With  any  given  set  of  coplanar  forces,  assume  the  addition 
of  any  force  whatever,  its  magnitude,  direction,  and  point  of 
application  being  chosen  entirely  at  random  or  to  suit  con- 


PARALLELOGRAM  OF  FORCES  AND  ITS  DERIVATIVES.        27 

P 

venience.  Find,  by  aid  of  the  magnitude  polygon,  the  mag- 
nitude and  direction  of  the  second  force  that  would  have  to  be 
added  to  the  set  to  close  the  magnitude  polygon  for  the  given 
set//z/.ythe  assumed  force.  This  second  force  would  be  the 
resultant  of  the  new  set,  and  if  properly  located  in  accordance 
with  the  method  of  §31,  would  be  a  force  which,  together 
with  the  assumed  force,  would  make  a  set  of  two  forces  which 
would  be  equivalent  to  or  balance  the  given  set  according  to 
the  senses  ascribed  to  them. 

The  senses  of  the  assumed  force  and  its  mate  will  be  con- 
fluent in  the  magnitude  polygon,  and  if  they  are  also  confluent 
with  the  rest  of  the  set,  the  two  would  balance  or  equilibrate 
the  given  set ;  if  they  are  not  so  confluent,  they  would  be  two 
forces  together  equivalent  to  the  given  set. 

Thus  two  forces  can  always  be  determined,  which  will  be 
equivalent  to  any  given  force  or  to  any  given  set  of  coplanar 
forces,  no  matter  how  complicated,  and  they  will  furnish  the 
desired  means  for  taking  account  of  points  of  application. 

It  may  be  observed  that  these  two  forces  must  always  fall 
within  one  of  five  cases  which  may  be  stated  and  interpreted  as 
follows : 

(a)  The  two  forces  may  be  inclined  to  each  other  and  either 
equal  or  unequal  in  magnitude.  By  the  parallelogram  of  forces 
their  resultant  must  pass  through  their  point  of  intersection. 
But  their  resultant  and  that  of  the  given  set  are  identical. 
Therefore  their  intersection  is  a  point  on  the  line  of  action  of 
the  resultant  of  the  given  set.  Their  non-parallelism  would 
show  that  there  must  be  such  a  resultant. 

Observe  that  this  method  furnishes  a  means  for  locating  the 
resultant  of  a  set  of  parallel  forces. 

(a'}  The  two  forces  may  be  parallel,  unequal,  and  opposite, 
proving  the  given  set  to  be  reducible  to  a  single  force  parallel 
to  the  two  forces  and  acting  at  a  point  determinate  only  by  re- 


28  STATICS. 

placing  the  two  forces  in  their  turn  by  two  others  equivalent  to 
them  and  inclined  to  each  other  as  in  (a).  This  may  be  re- 
garded as  a  special  case  of  (a).  This  case  can  always  be 
avoided  in  practice  by  judicious  selection  of  the  direction  of  the 
assumed  force. 

{a"}  The  two  forces  may  be  coincident,  unequal ',  and  opposite, 
proving  the  given  set  to  be  reducible  to  a  single  force,  itself 
coincident  with  the  two  coincident  forces.  This  like  (af)  may 
be  regarded  as  a  special  case  of  (a)  and  like  (a'}  can  always 
be  avoided  by  proper  selection  of  the  assumed  force. 

(b°)  The  two  forces  may  be  parallel,  equal,  and  opposite, 
forming  a  couple,  and  proving  the  given  set  to  be  reducible  to 
a  couple. 

(c)  The  two  forces  may  be  coincident,  equal,  and  opposite, 
that  is  in  equilibrium,  and  proving  the  given  set  to  be  in 
equilibrium. 

The  cases  (a),  (a'),  and  (a")  are  evidently  cases  in  which 
the  magnitude  polygon  is  not  closed,  and  (^)  and  (^)  are  evi- 
dently cases  in  which  that  polygon  is  closed. 

In  Figs.  6a,  6b,  6c  are  illustrated  the  important  cases  (a), 
(£),  and  (c)  respectively.  OA  and  oa  are  assumed  outright  and 
EO  and  eo  or  FO  and  fo  determined  just  as  AE  and  ae  were 
determined  in  Fig.  5.  In  Fig.  6a  it  appears  that  the  given  set 
amounts  to  a  single  force  AE\  in  Fig.  6b  to  an  anti-clockwise 
(negative)  couple  of  value  OA  x/;  and  in  Fig.  6c  that  they  are 
in  equilibrium,  A  and  F  coinciding  and  also  ao  and  af.  In 
Fig.  6a  the  case  (a')  would  have  resulted  if  O  had  been  taken 
anywhere  on  AE\  oa  and  oe  would  then  of  course  be  parallel, 
becoming  coincident,  case  (a"},  if  oa  should  chance  to  contain 
the  point  common  to  ab  and  ae. 

Observe  that,  in  Fig.  6a,  oa  and  oe,  and  in  Figs.  6b  and  6c, 
oa  and  cf,  with  the  senses  there  shown,  are  pairs  of  forces 
equivalent  to  the  given  set. 


PARALLELOGRAM  OF  FORCES  AND  ITS  DERIVATIVES.        29 

0 
The  forces  oa,  ob,  oc,  od,  etc.,  form  a  polygon  consisting 

of  the  lines  of  action  of  an  arbitrarily  chosen  force  oa,  and  the 


SCALE  OF  FORCE  MAGNITUDES 

0  20  40 


FIG.  6. 


resultants  respectively  of  oa  and  ab,  of  oa,  ab,  and  be,  etc., 
through  the  series  of  given  forces.  This  polygon  is  called  the 
string  polygon,  because  it  bears  a  certain  relation  to  the  form 


30  STATICS. 

which  a  string  would  assume  if  secured  at  the  ends  and  sub- 
jected to  the  given  forces.* 

If  the  two  added  forces  equivalent  to  or  balancing  the  given 
set  coincide,  the  string  polygon  forms  a  closed  figure  and  is 
described  as  closed. 

Obviously,  then,  if  the  string  polygon  closes,  the  given  set 
of  forces  cannot  be  reducible  to  a  couple. 

Observe  that,  in  Fig.  6a,  ae  was  located  where  ao  equal 
and  opposite  to  oa  would  have  given  rise  to  a  closed  string 
polygon. 

33.  Additional  Remarks  on  the  String  Polygon. — The 
sides  of  the  string  polygon  are  called  strings.  The  added  force, 
OA,  and  the  partial  resultants,  OB,  OC,  OD,  etc.,  form  a  set 
of  lines  radiating  from  a  common  point  (called  the  pole)  and 
are  hence  known  as  rays.  The  set  of  rays  and  the  magnitude 
polygon  constitute  the  whole  of  the  magnitude  diagram. 

To  construct  a  closed  string  polygon,  one  need  only  to  see 
that  the  strings  form  a  closed  polygon  having  an  apex  on  the 
line  of  action  of  each  and  every  force  of  the  system  (and  no- 
where else), — each  apex  to  be  formed  by  the  pair  of  strings 
which  represent  components  of  the  force  on  whose  line  of 
action  the  apex  appears. 

Any  one  string  is,  of  course,  a  closing  line  of  a  closed  string 
polygon,  but  any  string  which  it  may  be  convenient  to  draw 
last  will  commonly  be  spoken  of  as  the  closing  line. 

*  Terms  more  strictly  analogous  to  magnitude  polygon  would  be  point-of- 
application  polygon  or  location  polygon  but  they  are  not  in  use.  Funicular  polygon 
and  equilibrium  polygon  are  terms  in  common  use  as  well  as  string  polygon, 
the  former  being  merely  the  Latin  equivalent  to  string  polygon,  and  the  latter 
open  to  the  objection  that  it  might  with  equal  appropriateness  be  applied  to  the 
magnitude  polygon. 


CHAPTER    IV. 

ALGEBRAIC  AND  GRAPHIC  STATEMENTS  OF  THE  CONDI- 
TIONS OF  EQUILIBRIUM  WITH  APPLICATIONS. 

34.  Statical    Problems. — Problems    in    statics    deal    with 
bodies  which  may  be  conceived  to  be  at  rest  under  the  action 
of  a  set  of  forces,  some  of  which  are  not  fully  known,  i.e.,  not 
known  as  to  all  their  elements.      The  solution  of  such  prob- 
lems consists  of  finding  what  must  be  the  value  of  each  of  the 
unknown  elements.      This  at  once  suggests  the  use  of  algebra, 
and  the  first  and  principal  task  is  to  find  how  to  write  equations 
which  will  truly  represent  the  conditions — that  is,  which  will 
be  true  only  if  equilibrium  is  established.      The  forces  can  be 
represented  in  such  equations  only  by  expressions  for  their  ele- 
ments.     Such  elements  as  are  unknown  need  only  be  expressed 
as  unknowns,  to  be  evaluated  as  usual  by  the  ordinary  process 
of  solving  the  equations.      To  write  such  equations,  one  need 
only  to  consider  carefully  what  must  be  true  that  a  body  may 
be  at  rest,  i.e.,  what  the  conditions  really  are  which  are  to  be 
expressed. 

35.  Conditions  of  Equilibrium. — As  has  been  pointed  out 
(§  13),  conditions  of  rest  are  always  conditions  of  equilibrium, 
and  to  arrive  at  the  conditions  of  equilibrium  it  is  only  necessary 
to  consider  the  conditions  of  rest. 

In  order  that  a  body  may  be  at  rest,  two  things  must  be 
true  of  the  forces  acting  on  it,  viz. : 

(a)  The  total  tendency  of  some  of  them  to  produce  trans- 
Si 


32  STATICS. 

lation  in  one  direction  must  be  met  by  an  equal  total  tendency 
on  the  part  of  the  others  in  the  opposite  direction. 

(^)  The  total  tendency  of  any  of  them  to  produce  rotation 
in  one  direction  must  be  met  by  an  equal  total  tendency  on  the 
part  of  the  others  in  the  opposite  direction. 

Accordingly  a  set  of  forces  will  be  in  equilibrium  if  they 
can  result  in 

(A)  Neither  translation 

(B)  Nor  rotation. 

Observe  that  this  is  the  same  thing  as  saying  that  there 
must  be  for  a  resultant  neither  (A)  a  single  force  nor  (B)  a 
couple. 

It  remains  to  establish  the  algebraic  equivalent  of  (A)  and 
(^),  and  then  the  graphic  equivalent.  It  will  be  best  for  the 
present  to  confine  attention  to  coplanar  forces. 

36.  Algebraic  Statement  of  (A)  and  (B)  for  Coplanar 
Forces. — Since,  in  a  given  problem,  the  directions  of  the  forces 
may  be  very  various,  it  is  necessary  in  algebraic  work  to  sub- 
divide translation  in  general  into  two  component  translations 
parallel  to  a  set  of  axes,  preferably  rectangular.  For  coplanar 
forces  a  pair  of  such  axes  will  of  course  suffice.  They  will  be 
most  conveniently  taken  as  horizontal  and  vertical  and  will  be 
referred  to  as  the  axis  of  X  and  the  axis  of  Y  respectively,  ac- 
cording to  the  usual  convention. 

(A)  will  then  be  subdivided  into  two  partial  statements 
which  will  be  distinguished  by  the  proper  subscripts,  and  the 
conditions  can  now  be  written 

(Ax)  No  translation  to  the  right  or  left ; 
(Ay)  No  translation  up  or  down ; 
(B}  No  rotation. 

But  the  tendency  of  a  force  to  produce  translation  to  the 
right  or  left  is  simply  the  horizontal  or  Jf-component  of  the  force 
and  can  be  expressed  algebraically  Pl  cos  av  P2  cos  a2,  etc. 


CONDITIONS  OF  EQUILIBRIUM.  33 

•j 
And  the  tendency  of  a  force  to  produce  translation  up  or 

down  is  simply  the  vertical  or  F-component  of  the  force,  and  is 
expressed  algebraically  Pl  sin  av  P2  sin  #2,  etc. 

In  order  to  express  (B}  algebraically,  it  will  be  convenient 
to  use  the  principle  of  §  20  and  consider  the  given  set  replaced 
by  an  equivalent  set  consisting  of  a  series  of  concurrent  forces 
a~nd  a  series  of  couples.  The  point  of  concurrence  may  be 
chosen  to  suit  convenience  and  be  called  m.  The  couples 
need  to  be  considered  further  only  in  connection  with  rotation. 
The  moment  of  each  may  be  regarded  as  its  contribution 
towards  rotation,  and  the  values  of  these  moments  are  P^av 
P2a2,  P^,  etc.  If  now  m  be  taken  as  the  origin  of  coordi- 
nates, and  (xv  ^/j),  etc.,  be  taken  as  coordinates  of  the  points 
of  application  of  the  forces,  it  will  be  convenient  to  insert  the 
equivalents  for  the  a's,  and  the  moments  will  appear  in  the  form 
(P1j/1  cos  ^  —  P^XI  sin  «1),  (P2y2  cos  &2  —  P^2  sin  a2),  etc., 
\\hich  may  be  looked  upon  indifferently  as  the  sum  of  the 
moments  of  the  components  or  as  direct  substitution  of  values 
of  the  as.  Cf.  Cor.  II,  §  29. 

(A)  and  (B)  become,  accordingly,  expressed  as  equations: 

CPl  cos  ^  +  P2  cos  a2  +  .  .  .  +  Pn  cos  an  =  o.    .      A'x') 
(A")  \ 

lPl  sin  ai  +  P2  sin  a2  +  .  .  .  +  Pn  sin  an  =  O.    .      A'y') 

(B")     P,(^  cos  ai  -  ^  sin  «,)+... 

+  Pn(^n  COS  <*n  —  *H  sin  "n)    =  °» 

or,  more  briefly, 

(2Psin<x=o 
(Bf)  2P(y  cos  a  —  x  sin  a\  =  O, 


34  STATICS. 

or,  representing  all  horizontal  and  vertical  components  by  X 
and  Y  respectively,  these  three  equations  take  the  form  2X=  o, 
^2Y  =  o,  and  ^>(Xy  —  Yx)  =  o,  or  most  simply  of  all,  using 
Mfor  moments, 


2Y=o  .......     A,) 


and 

=  o 


This  last  is  the  usual  form  for  the  algebraic  statement  of 
conditions  of  equilibrium.  These  equations  form  an  equipment 
for  solving  statical  problems  algebraically. 

Two  other  ways  of  insuring  the  satisfaction  of  (A)  and  (B) 
algebraically  can  be  deduced,  one  of  much  practical  impor- 
tance, the  other  of  little.  These  will  receive  attention  in  due 
time  (§§  40,  41,  43),  and  will  be  shown  to  be  mere  transforma- 
tions of  the  equations  above  stated. 

Note  that  the  satisfaction  of  (A)  precludes  translation,  and 
that  the  satisfaction  of  (B)  precludes  rotation. 


Show  that  the  satisfaction  of  any  one  or  any  two  of  the 
Ax,  AyJ  and  B  equations  will  not  insure  equilibrium. 

Show  that  in  case  of  concurrent  forces  (A)  alone  suffices, 
and  gives  the  only  two  independent  equations  bearing  on  the 
case,  i.e.,  that  (B)  will  always  hold  in  this  Case  if  (A)  does. 

When  then  are  both  (A)  and  (B)  always  needed? 

Must  three  forces  in  equilibrium  be  concurrent  ?  Why  ? 
How  about  three  parallel  forces  ?  How  about  more  than  three 
forces  ?  Cf.  Cor.  I,  §  29. 

37.  Graphic  Interpretation  of  (A)  and  (B). — That  there 
may  be  no  translation  from  a  given  set  of  forces  it  is  evident 
that  their  resultant  must  be  zero,  and  by  §  3 1  it  is  seen  that 


CONDITIONS   OF  EQUILIBRIUM.  35 

• 
this  can  be  only  when  the  magnitude  polygon  for  the  set  of 

forces  closes.      We  can  state  then 

(A)  A  magnitude  polygon  must  close. 

That  there  may  be  no  rotation,  the  given  set  of  forces  must 
not  be  reducible  to  a  couple,  and,  as  has  been  shown  in  §  32, 
this  condition  will  be  fulfilled  if  a  string  polygon  closes  for  the 
set  of  forces.  We  can  state  accordingly 

(B°)  A  string  polygon  must  close. 

The  closed  magnitude  polygon  precludes  translation,  and 
the  closed  string  polygon  precludes  rotation.  These  two  poly- 
gons constitute  the  full  equipment  for  the  solution  of  statical 
problems  graphically.  It  remains  only  to  get  practice  in  using 
them. 

Here  again,  of  course,  for  concurrent  forces  (A)  alone 
suffices,  and  its  satisfaction  carries  with  it  the  certainty  of  sat- 
isfaction of  (B\  Hence  with  concurrent  forces  (j5)  is  super- 
fluous. 

38.  Exercises  in  the  Composition  of  Forces. — The  equip- 
ment for  solving  statical  problems  is  now  complete,  except  the 
transformation  of  the  Ax,  Ay,  and  B  equations  referred  to  in 
§  36,  and  developed  in  §§  40,41.  Everything  is,  however, 
entirely  ready  for  practice  in  the  composition  of  forces,  and 
practice  in  the  use  of  what  has  already  been  developed  will 
throw  much  light  on  what  is  to  follow. 

Below  are  given  explanations  of  the  algebraic  and  graphic 
methods  of  procedure  in  a  general  case.  In  special  cases  more 
or  less  of  both  processes  become  superfluous  and  can  be  omit- 
ted after  a  little  practice.  The  plan  for  arrangement  of  work 
and  for  retention  of  memoranda  at  all  steps  is  recommended 
as,  in  the  long  run,  an  economizer  of  time,  both  because  it  di- 


36  STATICS. 

minishes  the  risk  of  numerical  error,  and  because  it  aids  in  the 
location  of  such  error  when  once  committed. 

Let  it  be  required  to  determine  the  equilibrant  (and  result- 
ant) of  any  set  (say  five  in  number)  of  non-parallel,  non- 
concurrent  coplanar  forces. 

Graphic  Solution.  (Figs.  7 a,  ?b.  Cf.  PL  I.)  Letter  the 
five  forces  in  the  given  set,  preferably  in  the  order  of  their 
occurrence  from  left  to  right,  ab,  .  .  .  ,  de.  The  required 
equilibrant  will  be  ea  and  the  resultant  ae. 

Construct  (Fig.  'jb)  the  magnitude  polygon  (§31)  ABODE 
for  the  given  forces,  the  sides  having  lengths  proportional  to 
the  magnitudes  of  the  forces  to  which  they  are  parallel  and  so 
arranged  that  the  arrows  will  follow  one  another.  Evidently 
all  that  is  needed  to  close  this  polygon  now  is  to  fill  in  the 
side  EA.  EA  determines  the  magnitude  and  direction  of  the 
equilibrant,  and  AE  those  of  the  resultant. 

Construct  the  string  polygon  (Fig.  Jo).  The  question 
now  is  how  to  locate  ea  so  as  to  preclude  rotation,  i.e.,  how 
to  locate  ea  so  that  the  string  polygon  will  close  when  this 
force  is  added  to  the  set. 

Now  ea  must  act  at  the  intersection  of  strings  oe  and  ao, 
and  to  close  the  string  polygon  ao  must  coincide  with  oa. 
The  position  of  ao  thus  being  known  and  oe  having  already 
been  located  in  the  construction  of  the  string  polygon  up  to 
this  point,  it  only  remains  to  draw  ea  through  their  point  of 
intersection.  The  string  polygon  is  then  seen  to  be  closed, 
and  any  convenient  point  on  ae  may  be  taken  as  the  required 
point  of  application.* 

Algebraic  Solution.      (Cf.  PL  I.)     Let  the  unknown  magni- 


*  The  short  dash-lines  in  Fig.  "ja  show  ea  in  other  positions  than  the  correct 
one  and  the  positions  which  ao  would  in  those  cases  take,  leaving  the  string 
polygon  unclosed.  Such  forces  ea  and  ao  are  there  distinguished  by  subscripts. 


CONDITIONS  OF  EQUILIBRIUM. 


37 


tude,   direction,  and  point  of  application  be  designated  E,  a, 
and  (x,  y]  respectively.      Write  the  equations  for  horizontal 


(6) 


FIG.  7. 

and  vertical  components — the  Ax  and  Ay  equations — the  sec- 
ond considerably  below  the  first. 

Insert  in  small  figures,  as  memoranda,  the  values  of  the 


3^  STATICS. 

trigonometric  functions  just  above  each,  taking  care,  by  a 
moment's  inspection,  not  to  interchange  the  values  of  the  sines 
and  cosines. 

Perform  the  multiplications  indicated  and  insert  the  prod- 
ucts in  small  figures  as  memoranda  below  each  term. 

Proceed  with  the  reduction  as  shown  in  PL  I,  and  deter- 
mine the  values  of  E  and  a,  observing  that  the  a  found  will  be 
that  for  the  equilibrant  and  needs  only  to  be  changed  by  180° 
to  give  the  a  for  the  resultant.  The  quadrant  indicated  for  a 
is  the  first,  second,  third,  or  fourth  according  as  E  sin  a  and 
E  cos  a  are  both  plus,  the  former  plus  and  the  latter  minus, 
both  minus,  or  the  former  minus  and  the  latter  plus. 

Write  the  equation  of  moments — the  B  equation — in  ex- 
tended form,  inserting  the  values  of  the  given  coordinates  at 
once,  striking  out  all  terms  in  which  a  zero  coordinate  appears. 

Write  as  memoranda  above  each  remaining  P  cos  a  and 
P  sin  a  the  value  found  for  it  in  reducing  the  two  preceding 
equations. 

Perform  the  multiplications  indicated  and  show  products 
under  each  term  as  memoranda,  and  solve,  assuming  the  value 
for  either  x  or  y  as  may  be  more  convenient. 


Specifications  for  Exercises  3-7.  In  each  exercise  the  object  is  to  work  out 
both  graphically  and  algebraically,  the  equilibrant  and  resultant  of  the  sets  of  co- 
planar  forces  there  given.  The  results  should  be  recorded  side  by  side  conspic- 
uously for  comparison.  Work  will  be  arranged  in  general  as  in  PI.  I,  and  each 
exercise  should  be  undertaken  as  a  general  case  and  so  carried  to  completion. 

Magnitudes  are  expressed  in  pounds  and  lengths  in  feet. 

Angles  will  best  be  laid  off  by  plotting  from  a  table  of  natural  tangents,  and 
in  favorable  cases  by  the  use  of  triangles  and  T-square.  Small  protractors  will 
sometimes  be  found  useful  for  checking. 

For  scales,  I  in.  =  20  Ibs.  and  I  in.  =  4  or  5  feet  are  recommended. 

Exercise  3.  Non-concurrent,  non-parallel  forces :  (20  105°  5,  o), 
(25  263°  ii,  o),  (30  98°  15,  o),  (10  80°  17,  o),  (15  280°  2,  o). 

Exercise  4.  Parallel  forces  uniform  in  sense;  five  vertically  upward 
forces  of  10,  20,  30,  40,  and  50  Ibs.  respectively,  with  successive  inter- 
vals between  them  of  3,  2,  6  and  5  ft. 


CONDITIONS  OF  EQUILIBRIUM.  39 

P 

Exercise  5.  Parallel  forces  varying  in  sense :  the  forces  of  the  pre- 
ceding exercise  but  with  the  lo-lb.  and  3o-lb.  forces  reversed  in  sense. 

Exercise  6.  Concurrent  forces  :  five  forces  of  10,  20,  30,  40,  and  50  Ibs. 
respectively,  with  directions  of  30°,  60°,  135°,  220°,  and  320°  all  applied  at 
(o,  o). 

Exercise  7.  Three  non-concurrent  forces  whose  magnitude  polygon 
closes:  (50  330°  —  6,  o),  (50  210°  o,  2),  and  (50  90°  8,  o). 

NOTE. — This  is  a  case  similar  to  that  of  Fig.  6£,  §  32.  The  results  can  be 
stated  in  foot-pounds  with  the  proper  algebraic  sign  to  indicate  sense,  say  -(-  for 
clockwise  and  —  for  anti-clockwise  rotation.  The  source  of  graphical  results 
should  be  made  clear  by  use  of  dimension  lines,  including  scaled  numerical  re- 
sults and  the  expression  of  the  necessary  multiplication.  There  is  no  reason 
why  OA  may  not  be  taken  of  some  round  magnitude,  say  20,  30,  etc.,  and  it  will 
be  advantageous  in  this  case. 

39.  Generalization  of  the  Three  Classes  of  Resultants. — 

It  may  be  observed  in  passing  that  the  three  classes  of  result- 
ants to  one  of  which  any  set  of  coplanar  forces  may  be  re- 
duced can  be  expressed  in  generalized  form  in  the  notation 
of  this  book,  as  follows : 

Class  i.  (>o,  o°  to  360°,  unrestricted),  i.e.,  a  single  force. 
In  this  case  the  magnitude  polygon  fails  to  close,  and  the  string 
polygon  may  or  may  not  close. 

Class  2.  (o,  indeterminate,  at  infinity),  i.e.,  a  couple. 
Here  the  magnitude  polygon  closes,  but  the  string  polygon 
does  not  close. 

Class  3.  (o,  indeterminate,  indeterminate),  i.e.,  a  set  in 
equilibrium.  The  magnitude  polygon  and  the  string  polygon 
both  close. 

Examples  of  the  first  two  of  these  classes  are  contained  in 
the  data  for  Exercises  3-7.  Inserting  the  equilibrant  there 
required,  the  class  then  exemplified  is  in  each  case  the  last  one, 
of  course. 

40.  Establishment  of  Equilibrium   by  Use  of  Moments 
Alone. — While  the  algebraic  statement  of  equilibrium  of  §  36 
is,  of  course,  of  universal  validity  in  cases  of  coplanar  forces, 
there  is  an  alternative  algebraic  statement  which  in  many  cases 


40  STATICS. 

proves  much  more  convenient  in  use.  This  statement  takes 
into  account  moments  only,  and  amounts  simply  to  three 
equations  of  moments  with  certain  limitations  laid  upon  the 
choice  of  the  centers  of  moments.  This  alternative  is  deduced 
as  follows. 

To  establish  equilibrium,  it  must  be  made  certain  that  the 
given  S£t  of  forces  will  cause  neither  translation  nor  rotation, 
i.e.,  that  they  amount  to  neither  a  single  force  nor  a  couple. 
It  has  been  shown  that  there  can  be  no  couple  if  2M —  o  for 
any  point  whatever.  It  remains  to  be  shown  how  moments 
can  be  used  to  prove  the  non-existence  of  a  single  force  re- 
sultant as  well. 

Suppose  that  for  a  given  center  of  moments  2M  =o  for  a 
given  set  of  forces.  This  can  arise  from  only  one  of  two 
causes :  either  the  set  of  forces  is  in  equilibrium,  and  would 
have  no  moment  about  any  point,  or  the  center  of  moments  lies 
upon  the  resultant  of  the  set.  If,  then,  it  can  be  shown  that 
the  center  of  moments  does  not  lie  on  the  resultant  of  the  set, 
the  set  must  be  in  equilibrium. 

Any  one  or  any  two  points  chosen  at  random  might  both 
chance  to  lie  on  the  line  of  action  of  a  resultant.  In  such  a 
case  12M  would  be  zero  in  spite  of  the  resultant  having  a  mag- 
nitude greater  than  zero,  and  if  investigation  went  no  further 
translation  might  exist  undetected.  If,  however,  three  points 
not  in  the  saine  straight  line  be  chosen,  one  of  the  points  at 
least  will  not  lie  on  such  line  of  action.  If  2M  =  o  for  each 
of  these  points,  2M  =  o  for  at  least  one  point  not  on  the  line 
of  action  of  the  resultant,  and  equilibrium  is  a  certainty. 
Hence  it  can  be  stated  that 

If,  for  a  set  of  forces,  2M  is  zero  for  three  points  not  in 
the  same  straight  line,  equilibrium  is  assured. 

It  should  be  carefully  observed  that  the  three  equations  of 
moments  implied  in  the  preceding  sentence  are  merely  the  Ax, 


CONDITIONS  OF  EQUILIBRIUM.  41 

d 

A  ,  and  B  equations  themselves,  though  transformed  *  in  such 
a  way  as  to  be  more  convenient  for  use  in  certain  problems. 
As  mere  variants  or  alternative  forms  of  the  equations  of  equi- 
librium first  deduced  they  can  solve  no  problems  which  the 
former  could  not  be  made  to  solve,  and  any  general  conclusions 
drawn  from  inspection  of  either  of  the  two  sets  of  equations 
could  be  drawn  also  from  inspection  of  the  other.  The  Ax, 
Ayt  and  B  equations,  as  they  are  of  simpler  algebraic  form,  are 
more  convenient  for  such  inspection,  but,  nevertheless,  they 
are  a  group  of  equations  algebraically  identical  with  the  three 
moment  equations. 

41.  Determination  of  Magnitudes  by  Single  Moment 
Equations  Alone. — Complete  determination  of  equilibrium 
requires  in  general  three  equations,  AX9  Ay,  and  B,  or  an  equiva- 
lent, such  as  the  three  moment  equations  of  §  40.  Still  it  will 
be  worth  while  to  see  under  what  circumstances,  if  at  all,  a 
single  equation  of  moments  will  determine  the  magnitude  of  any 
one  of  three  or  fewer  forces  whose  lines  of  action  are  known. 

Call  this  unknown  force  <2»  and  the  others  R  and  5. 
Then,  since  Q  is  to  be  the  equilibrant  of  all  the  rest  of  the  set, 
its  line  of  action  must  be  that  of  the  resultant  of  the  set. 


*  The  nature  of  this  transformation  appears  upon  writing  one  of  these  moment 
equations  in  its  general  form  as  follows. 

If  the  points  of  application  of  the  forces  be  a  series  of  (AT,  y)'s,  and  any  one  of 
the  centers  have  the  coordinates  (xa,  ya)  with  reference  to  the  same  axes,  the 
moment  equation  will  take  the  form 


-ya)-      *  -*.)    =  o; 

or,  noting  that  xa  and  ya  are  constants  and  all  the  rest  of  the  letters  variables, 
2(Xy  -  Kr)  -  yaSX  +  xa2  Y  =  o, 

which  is  obviously  B  minus  Ax,  multiplied  through  by  a  constant  and  plus  Ay 
multiplied  through  by  another  constant.  One  or  both  of  these  constants  might, 
of  course,  be  zero.  If  the  three  centers  be  taken  at  the  origin  and  elsewhere  on 
each  of  the  axes  of  x  and  y  respectively,  the  three  moment  equations  become 
simply  B,  and  B  —  Ax  X  a  constant  and  B  -j-  Ay  X  a  constant. 


42  STATICS. 

Realizing  this  it  is  easy  to  select  a  center  which  certainly  does 
not  lie  on  the  line  of  action  of  the  resultant  of  the  set.  R 
and  5,  as  well  as  Qy  will  in  general  produce  moments  about  the 
point,  and,  appearing  in  the  equation,  will  prevent  Q  from 
being  the  only  unknown  in  it.  If,  however,  the  point  can  be 
selected  so  that  it  will  be  off  the  line  of  Q,  and  at  the  same 
time  cause  the  arms  of  R  and  S,  or  of  5  if  5  is  the  only  other 
unknown,  to  be  zero,  determination  of  Q  by  one  equation  is 
possible.  Such  a  point  is  to  be  found  at  the  intersection  of  R 
and  St  or  anywhere  on  5  if  5  is  the  only  other  unknown. 
Cases  in  which  this  cannot  be  done  arise  only  when  there  are 
more  than  three  unknowns,  or  when  the  lines  of  action  of  three 
are  parallel  or  concurrent. 

An  equation  thus  derived  of  course  does  not  in  general 
establish  equilibrium,  for  the  establishment  of  equilibrium  re- 
quires the  determination  of  two  or  three  elements,  and  for  this 
more  than  one  equation  is  always  necessary ;  but  it  does  deter- 
mine one  magnitude  to  which  the  other  forces  must  inevitably 
conform  in  order  to  produce  equilibrium.  To  get  more,  simply 
repeat  the  process.  In  general,  then, 

To  find  the  magnitude  of  a  force  whose  line  of  action 
is  given,  solve  an  equation  of  moments  for  a  center  on  the 
line  of  action  of  the  other  force  whose  magnitude  is  un- 
known ;  or,  if  there  are  two  others,  for  a  center  at  their 
intersection.  Repeat  the  process  for  the  other  one  or  two 
forces. 

This  method  is  of  great  use  whenever  the  magnitude  is 
sought  for  a  force  with  a  given  line  of  action.  Such  problems 
are  very  common  in  practice.  In  such  rases  the  method  is 
often  useful  for  determining  senses  of  forces  by  mere  inspec- 
tion without  any  calculation. 

Notice  that  solving  Case  4  in  this  way  is  simply  utilizing 


CONDITIONS  OF  EQUILIBRIUM.  43 

the  general  principles  of  the  preceding  section,  choosing  the 
centers  so  as  to  avoid  simultaneous  equations. 

The  amount  of  it  is  that  if  there  is  equilibrium,  2M  =  o 
for  any  center,  but  the  converse  is  not  true.  Hence  the  need 
of  more  than  one  moment  equation. 

Suppose  the  condition  requiring  the  center  to  be  off  the 
line  of  action  of  <2  were  violated,  how  would  the  equation  be 
affected  ? 

42.  Convention  as  to  Algebraic  Signs  in  Moment  Equa- 
tions.— In  using  Ax,  Ay,  and  B  it  has  been  convenient  to  con- 
sider magnitudes  of  forces  as  always  positive  and  to  associate 
sense  exclusively  with   slope.      In   using  the  method  of  mo- 
ments it  will  be  more  convenient  to  call  arms  universally  posi- 
tive, and  give  magnitudes  varying  signs :  -f-  if  for  a  given  cen- 
ter they  cause  rotations  in  one  sense,  and  —  if  the  opposite. 
Throughout    these    pages    clockwise  rotations  will  be    called 
positive,  and  anti-clockwise  negative. 

To  apply  the  method  of  moments,  select  the  center,  write 
the  equation  calling  all  moments  positive  that  are  not  known 
to  be  negative.  The  sign  found  with  the  result  will,  in  con- 
nection with  the  center  for  the  equation,  determine  the  sense 
of  the  force.  The  same  force  may  be  positive  for  one  center 
and  negative  for  another,  hence  the  locations  of  the  centers 
should  be  recorded  near  the  work. 

43.  Six  Methods  of  Stating  the  Conditions  of  Equilib- 
rium.— In  addition  to  the  statements  of  the  conditions  of  equi- 
librium already  given,  there  is  a  third  from  the  algebraic  and 
two  more  from  the  graphic  point  of  view.     The  six  ways  are 
here  collected  and  stated  for  completeness,  and  for  reference: 


44 


STATICS. 


Graphic. 
Equilibrium  will  exist  if 

(1)  A  string  polygon  closes 
for  each  of  three  poles  not  in 
the  same  straight  line ; 

or  if 

(2)  Two     string     polygons 
close  with  the  same  pole ; 

or  if 


Algebraic* 
Equilibrium  will  exist  if 

(1)  The   sum   of  the  mo- 
ments is    zero    for  each  of 
three  points  not  in  the  same 
straight  line; 

or  if 

(2)  The  sum  of  the  moments 
is  zero  for  each  of  two  points, 
and  the  sum  of  the  components 
is  zero  along  a  line  not  per- 
pendicular to  the  one  joining 
these  two  points; 

or  if 

(3)  The   sum  of  the  mo- 
ments is  zero  for  one  point 
and  the  sum  of  the  compo- 
nents is  zero  for  hoth  of  any 
two  directions. 

The  statements  shown  in  bold  face  are  the  ones  already 
developed  at  length  and  are  the  only  ones  of  practical  use. 
The  critical  reader,  however,  will  assure  himself  of  the  correct- 
ness of  the  three  others. 

*  The  first  of  these  statements  was  shown  in  §  40  to  be  an  algebraic  identity 
with  the  third,  and  deducible  directly  from  it,  and  the  same  might  be  proved 
of  the  second  in  a  similar  manner. 


(3)  One  string  polygon  and 
one  magnitude  polygon  close. 


CHAPTER   V. 
SCOPE   OF   PURE   STATICS. 

44.  General  Survey  of  the  Scope  of  Pure  Statics. — The 
term  Pure  Statics  is  to  be  understood  to  mean  the  perfectly 
general  science,  in  which  problems  can  have  no  light  thrown 
on  them  by  considerations  of  size,  shape,  and  elasticity  of 
bodies.  Its  only  resources  are  the  conditions  of  equilibrium  of 
rigid  bodies. 

The  first  algebraic  statement  of  these  conditions  which 
has  been  elaborated  is  of  such  a  form  as  specially  to  invite 
thought  as  to  the  scope  of  the  subject — as  to  how  many  and 
what  coplanar  statical  problems  are  capable  of  solution. 

First  may  be  considered 

(a)  Non-concurrent  Forces.  There  are  only  three  funda- 
mental equations  (§43),  Ax,  Ay,  and  B.  If,  then,  in  any 
problem  there  are  more  than  three  forces  of  which  elements 
are  not  known,  there  are  more  than  three  unknown  quantities 
and  the  problem  of  establishing  equilibrium  is  indeterminate. 
Hence  all  forces  but  three  (at  most)  must  be  fully  known.  But 
three  forces  involve  nine  elements ;  hence,  unless  six  of  these 
elements  are  known  the  problem  is  still  indeterminate. 

These  are  a  priori  considerations  regardless  of  the  struc- 
ture of  these  particular  equations.  Taking  this  structure  into 
account,  the  field  becomes  still  more  circumscribed.  For 
example,  since  only  one  of  the  equations  involves  points  of 

45 


46  STATICS. 

application,  it  is  clear  that  there  must  be  not  more  than  one 
such  point  unknown.  Even  then  the  point  is,  strictly  speak- 
ing, indeterminate,  for  the  point  appears  as  two  unknown 
quantities,  .*•  andj/.  For  the  purposes  of  statics,  however,  any 
pair  of  values  for  x  and  y  that  will  satisfy  the  equation  will 
meet  all  requirements,  and  this  actual  indetermination  is  not  a 
source  of  difficulty.  In  other  words,  B  becomes  simply  the 
equation  of  the  line  of  action  of  the  force  in  terms  of  x  and  y, 
on  which  an  indefinite  number  of  points  of  application  may  be 
selected. 

Nine  elements  can  be  divided  into  groups  of  six  known  and 
three  unknown  in  9x8x7-^-3X2  =  84  different  ways, 
which  reduce  to  twenty  separate  cases  at  once  (see  Appendix 
for  a  complete  statement  of  them),  and  of  these,  fifteen  are 
more  or  less  indeterminate,  owing  to  the  peculiar  limitations  of 
the  equations  above  pointed  out.  Moreover,  only  three  of 
the  twenty  cases  are  important.  They  are  the  following  : 

a.  When  the  three  unknown  elements  all  pertain  to  one 
force,  i.e.,  one  force  is  wholly  wanting.  This  is  the  problem 
of  finding  the  resultant  and  the  equilibrant  of  a  given  set  — 
Composition  of  Forces. 

j3.  When  the  three  unknown  elements  pertain  to  two  forces 
and  the  magnitude  of  one  and  the  magnitude  and  slope  of  the 
other  are  wanting. 

y.  When  the  three  elements  pertain  to  three  forces,  and 
three  magnitudes  are  wanting. 

There  are  still  to  be  considered 

(b)  Concurrent  (including  Parallel)  Forces.  Here  there 
are  only  two  fundamental  equations,  Ax  and  Ay*  one  of 


*  The  B  equation  becomes  simply  a  mathematical  identity  with  Ax  and  Av  in 
the  case  of  concurrent  forces,  for  the  x's  andjy's  in  ~2{Xy  —  Yx}=  o  are  then  each 
constant,  say  xa  and  ya.  The  equation  is  then  ya  2X  —  xa  2  Y  =  o,  which  is 


SCOPE  OF  PURE  STATICS.  47 

f 
which  must  be  replaced  by  B  if  the  forces  are  parallel.     They 

can  determine  only  two  unknown  quantities.  If  there  are 
more  than  two  forces  of  which  elements  are  unknown,  there  are 
more  than  two  quantities  unknown  and  the  problem  of  estab- 
lishing equilibrium  is  indeterminate.  Hence  all  forces  but  two 
(at  most)  in  a  set  of  concurrent  or  parallel  forces  must  be 
fully  known.  But  two  forces  involve  six  elements  which  may 
be  unknown,  and  unless  two  of  these  elements  are  known,  the 
problem  is  indeterminate.  Six  elements  may  be  divided  into 
groups  of  four  known  and  two  unknown  in6x  5  -5-  2  =  15 
different  ways,  which  reduce  to  nine  separate  problems  in  the 
equilibrium  of  concurrent  or  parallel  forces  (see  Appendix). 
Noting  that  composition  of  such  forces  is  included  in  the  gen- 
eral problem  of  composition  of  forces  (<*)  as  a  special  case, 
only  one  of  the  rest  of  the  nine  cases  is  important,  viz. : 

o\  When  the  two  elements  pertain  to  two  forces,  and  the 
magnitudes  of  the  two  forces  are  wanting, — Resolution  of 
Forces. 

45.  The  Four  Cases, — Only  four  cases  capable  of  solution 
by  pure  statics  are  of  sufficient  importance  to  require  special 
study.  These  have  been  pointed  out  in  §  44,  and  will  here  be 
restated  and  numbered  for  future  reference. 

Case  i.  Magnitude,  direction,  and  point  of  application  of 
one  force  required, — Composition  of  Forces. 

Case  2.  Magnitudes  of  two  forces  required, — Resolution  of 
Forces. 

This  case  will  be  divided  into  two  sub-cases,  2.0.  and  ib.  In  la  the 
two  forces  are  non-parallel  ;  in  ib,  parallel. 

Case  3.  Magnitude  of  one  force  and  magnitude  and  direc- 
tion of  another  required. 

Case  4.    Magnitudes  of  three  forces  required.* 

*  As  will  be  seen  (  §  5<3a),  Case  4  may  be  regarded  as  a  mere  combination  of 
Cases  3  and  2a,  and  Case  3  as  combination  of  Cases  4  and  I. 


48  STATICS. 

This  case  is  sometimes  spoken  of  as  Resolution  in  Three 
Directions. 

Case  4  is  statically  determinate  only  when  the  three  forces 
are  neither  concurrent  nor  parallel. 

All  these  four  cases  or  any  of  the  few  others  which  are 
capable  of  solution  can,  of  course,  be  solved  by  applying  Ax, 
Ayy  and  B — in  some  cases  more  conveniently  in  the  form  of 
the  moment  equations  of  §40  or  §41 — or  by  closing  the  mag- 
nitude and  string  polygons. 

See  Plates  I-V  respectively  for  a  full  treatment  of  these 
four  cases.  These  plates  are  explained  in  the  following 
chapter. 


CHAPTER    VI. 

SOLUTION  OF  STATICAL  PROBLEMS,  WITH  SPECIAL  REF- 
ERENCE TO  THE  FOUR  MOST  IMPORTANT  CASES. 

46.  The  Solution  of  Statical  Problems. — In  general  the 
first  thing  to  be  done  in  undertaking  a  problem  in  statics  is  to 
draw  a  sketch  showing  all  that  is  known  of  all  the  forces, — 
their  magnitudes,  either  as  knowns  or  unknowns,  being  in- 
scribed near  their  lines  of  action.  This  operation  is  of  funda- 
mental importance  and  is  the  step  of  greatest  practical 
difficulty.  It  is  sometimes  called  "  showing  the  forces."  It 
consists  really  of  accurately  analyzing  the  situation  and  decid- 
ing just  what  forces  are  acting  and  just  what  is  known  of  each. 
This  once  done,  and  the  problem  seen  to  be  determinate,  it  is 
a  matter  of  mere  routine  to  proceed  with  the  solution. 

The  mastery  of  these  solutions  is  what  study  of  this  chap- 
ter is  expected  to  bring  about.  The  ability  to  show  the  forces 
will  require  a  longer  time  to  cultivate,  and  its  importance  may 
well  be  kept  in  mind. 

The  methods  of  solution  for  the  four  important  cases  will 
now  be  taken  up.  It  is  well  to  note  at  the  outset  that  the  whole 
equipment  available  consists  of  the  magnitude  polygon  and  the 
string  polygon  for  graphic  solutions,  and  of  the  Ax,  Ay,  and 
B  equations,  or  their  alternative,  the  three  moment  equations  of 
§  40  for  algebraic  solutions.  The  task  is  now  simply  to  become 
familiar  with  the  manipulations  by  which  this  equipment  can 
be  made  to  fit  the  peculiarities  of  each  case,  and  to  yield  most 
easily  the  desired  results. 

49 


50  STATICS. 

In  all  cases,  it  should  be  noted,  it  will  be  found  most  con- 
venient in  graphical  work  to  letter  the  known  forces  in  the 
order  of  their  occurrence,  going  from  left  to  right  around  the 
body  on  which  they  act,  leaving  the  unknowns  to  be  lettered 
consecutively  after  the  knowns  have  all  been  provided  for. 
The  first  letter  will  naturally  be  a,  and  the  last  letter  of  the 
last  unknown  may  properly  be  a.  Thus  will  be  avoided  double 
letters  for  apices  of  the  magnitude  diagram. 

Furthermore  in  algebraic  work  senses  of  forces  given  only 
in  line  of  action  will  be  assumed  to  be  such  as  to  make  their 
directions  less  than  1 80°,  in  writing  the  Ax  and  Ay  equations, 
and  such  as  to  give  a  positive  moment  in  writing  the  non- 
simultaneous  moment  equations  required  for  some  of  these 
cases.  If  the  sense  so  assumed  is  the  correct  one,  the  fact 
will  appear  by  the  magnitudes  proving  to  be  plus  quantities. 
If  the  magnitudes  prove  negative,  of  course  the  correct  sense 
is  the  reverse  of  that  assumed. 

After  writing  an  equation  of  moments  it  will  be  well  to 
check  the  lengths  of  all  arms,  whether  given  or  not,  by  scal- 
ing them  from  a  drawing,  if  one  be  at  hand.  Check  also  so 
far  as  possible  the  determinations  of  senses  by  inspection  of  the 
drawing. 

47.  Solution  of  Case  I. — Case  I  is  mentioned  here  merely 
for  completeness,  and  the  account  of  its  solution  given  in  §  38, 
and  presumably  familiar  to  the  reader  by  this  time,  need  not 
be  reprinted  here. 

48.  Solution  of    Case   2a.     (Cf.  PI.  lla.} — Required  the 
magnitudes  (and  senses)  of  two  non-parallel  forces  in  a  set  in 
equilibrium. 

Algebraic  Solution.  Let  P  and  Q  be  the  unknown  mag- 
nitudes, and  assume  trial  senses  as  per  the  next  to  the  last 
paragraph  of  §  46. 

Solve,    for  P  and  <2,   either  (a)  the  Ax  and  Ay  equations 


SOLUTION  OF  STATICAL  PROBLEMS.  51 

§ 

(avoiding  simultaneity,  if  desired,  by  taking  one  axis  of  refer- 
ence along  one  of  the  unknowns),  or  (£)  their  variants,  the 
two  moment  equations  of  §  41.  The  assumption  as  to  sense 
made  at  the  outset  is  correct  or  must  be  reversed  for  either 
force  according  as  its  magnitude  comes  out  plus  or  minus. 

Graphic  Solution  (Fig.    8  and  PL  \\a).     Close  the   mag- 


FIG.  8. 

nitude  polygon  (Fig.  8£).  Suppose  the  first  known  of  the 
given  set  be  ab  and  the  two  unknowns  de  and  ea.  The  mag- 
nitude polygon  can  be  completed  by  familiar  means  from  the 
apex  A  to  D  inclusive.  Then  DE  drawn  parallel  to  de  and 
AE  parallel  to  ae  will  locate  the  missing  apex  E,  and  the  re- 
quired magnitudes  and  senses  will  be  DE  and  EA . 

The  string  polygon  is  superfluous.  It  is  bound  to  close  if 
the  magnitude  polygon  is  closed. 

NOTE. — Case  2a  nearly  always  occurs  in  practice  for  a  set  of  concurrent 
forces,  but  it  is  capable  of  solution  as  well  for  a  non- concur  rent  set  if  the  resultant 
of  all  the  knowns  passes  through  the  point  of  intersection  of  the  two  unknowns. 

48a.  Solution  of  Case  2b.  (Cf.  PL  lib.}— Required  the 
magnitudes  (and  senses)  of  two  parallel  forces  in  a  set  in 
equilibrium. 


STATICS. 


Algebraic  Solution.  Let  P  and  Q  be  the  unknown  mag- 
nitudes and  assume  trial  senses  as  per  the  next  fo  the  last 
paragraph  of  §  46. 

As  P  and  Q  are  parallel,  the  JT-axis  may  well  be  taken  at 
right  angles  to  them  both,  making  their  directions  90°. 

Write  the  Ax  and  Ay  equations  as  in  Case  2a.  Noting  that  the  Ax 
equation  vanishes,  its  place  is  made  good  by  the  ^equation  (retaining  the 
same  senses  as  in  the  preceding  equations)  written  as  in  Case  i.  Solving 
the  simultaneous  equations,  P  and  Q  are  known,  and  the  assumed  senses 
are  correct  or  to  be  reversed  according  as  results  come  out  plus  or 
minus. 

Or,  better,  solve  by  moments  alone  by  the  method  of  §  41, 
taking  a  center  on  the  line  of  action  of  P  to  find  Q,  and  on 


e 
P   ? 


(a) 


aa 


bb 


d  d 


FIG.  9. 

the  line  of  action  of  Q  to  find  P.  The  assumed  senses  are  cor- 
rect or  to  be  reversed  according  as  results  come  out  plus  or 
minus.  Check  results  by  seeing  whether  Ay  is  satisfied  finally. 


SOLUTION  OF  STATICAL  PROBLEMS.  53 


Graphic  Solution  (Fig.  9  and  PL  11^).  Let  the  knowns 
be  lettered  ab  .  .  .  cd,  and  the  unknowns  accordingly  de  and  ea. 

Construct  the  magnitude  polygon  from  A  to  D,  inclusive 
(Fig.  9<£).  Assume  a  convenient  oa  and  construct  the  string 
polygon  to  od,  inclusive  (Fig.  go).  In  order  that  ao  may  coin- 
cide with  oa  and  close  the  string  polygon,  oe,  the  missing 
string,  must  intersect  ea  where  oa  does,  and  oe  must  start 
from  the  intersection  of  od  and  de.  Hence  oe  is  located,  and 
the  string  polygon  is  closed. 

E  must  then  be  on  a  ray  from  O  parallel  to  oe,  and  on  a 
line  from  A  or  D  parallel  to  ae  or  de.  E  is  therefore  located, 
the  magnitude  polygon  is  closed,  and  the  required  magnitudes 
and  senses  will  be  DE  and  EA . 

NOTE. — This  case  almost  always  occurs  in  practice  for  a  set  of  parallel  forces, 
but  it  is  capable  of  solution  as  well  for  any  set  provided  that  the  resultant  of 
the  knowns  is  parallel  to  the  two  unknowns. 

49.  Solution  of  Case  3.  (Cf.  PL  III.) — Required  the 
magnitude  and  direction,  and  the  magnitude  (and  sense)  re- 
spectively, of  two  forces  in  a  set  in  equilibrium. 

Algebraic  Solution.  Let  P  and  Q  be  the  two  unknown 
magnitudes  and  a  the  direction  of  P.  Assume  the  sense  of 
Q  as  per  the  next  to  the  last  paragraph  of  §  46. 

Write  the  B  equation,  taking  the  center  on  the  line  of 
action  of  P,  and  establish  the  magnitude  and  sense  of  Q  (§  41). 

One  point  only  in  the  line  of  action  of  P  being  given,  that 
point  must,  of  course,  be  the  center  selected. 

In  writing  B,  memoranda  may  advantageously  be  inserted 
near  each  term,  showing  the  value  of  each  component  of  each 
force,  as  well  as  the  moment  of  each  component. 

Write  the  Ax  and  Av  equations,  inserting  the  value  of  <2just 
established,  and  get  the  horizontal  and  vertical  components  of 
P  and  establish  P  and  a  by  routine  like  that  for  a  similar  pur- 
pose in  Case  i . 


54 


STATICS. 


Observe  that  the  memoranda  inserted  near  the  terms  of  the 
B  equation  can  be  utilized  in  the  Ax  and  Ay  equations,  just  as 
the  reverse  was  done  in  Case  i. 

Observe  also  that  Q  once  fully  known,  the  problem  is  simply 
that  of  finding  an  equilibrant  whose  point  of  application  is 
known  in  advance,  and  the  work  is  thenceforth  that  of  Case  I 
for  finding  the  magnitude  and  direction  of  the  equilibrant. 

Graphic  Solution  (Fig.  10  and  PI.  III).  Let  the  knowns 
be  lettered  ab  .  .  .  cd,  and  the  unknowns  accordingly  de 


(5) 


FIG.  10. 

and  ed,  the  former  being  unknown  in  magnitude  (and  sense) 
and  the  latter  in  magnitude  and  direction. 

Construct  the  magnitude  polygon  from  A  to  Z>,  inclusive 
(Fig.  io£).  Assume  a  convenient  oa,  draw  it  through  the 
given  point  of  application  of  ea,  and  construct  the  string 
polygon  to  od,  inclusive  (Fig.  io<z).  In  order  that  ao  may  co- 
incide with  oa  and  close  the  string  polygon,  oe,  the  missing 
string,  must  intersect  ea  where  oa  does,  i.e.,  at  the  only  known 


SOLUTION  OF  STATICAL  PROBLEMS.  55 

point  of  ea ;  furthermore  oe  must  start  from  the  intersection  of 
od  and  de.  Hence  oe  is  located  and  the  string  polygon  is 
closed. 

E  must  then  lie  on  a  ray  from  O  parallel  to  oe>  and  on  a 
line  from  D  parallel  to  de.  E  is  thus  located,  and  connect- 
ing E  and  A,  the  magnitude  polygon  is  closed.  The  required 
magnitude  (and  sense),  and  magnitude  and  direction,  are  DP. 
and  EA  respectively. 

Observe  that  the  key  to  this  solution  lies  in  locating  oa  in 
the  only  way  possible,  so  that  its  point  of  intersection  with  ea 
may  be  known;  i.e.,  by  drawing  it  through  the  only  point 
through  which  ea  is  known  at  the  outset  to  pass. 

50.  Solution  of  Case  4.  (Cf.  Pis.  IV  and  V.)— Required 
the  magnitudes  (and  senses)  of  three  forces  in  a  set  in  equi- 
librium. 

Algebraic  Solution.  Let  P,  Q,  and  R  be  the  required  mag- 
nitudes. Assume  senses  as  per  the  next  to  the  last  paragraph 
of  §  46. 

The  Ax,  Ay  and  B  equations  can  now  be  written,  taking  the  center  for 
B  anywhere  whatever,  as  in  PI.  IV,  and  there  result  three  simultaneous 
equations,  whence  the  desired  quantities  can  be  evaluated.  This  method 
is  needlessly  laborious,  and  the  simultaneous  equations  can  always  be 
avoided  by  the  aid  of  the  three  moment  equations  of  §  41,  as  follows. 

Write  three  equations  of  moments,  taking  the  centers  at  the 
intersections  respectively  of  Q  and  R,  R  and  P,  and'  P  and  Q. 
These  equations  will  yield  the  magnitudes  (and  senses)  of  P, 
Q,  and  R  respectively. 

Obviously  the  solution  is  indeterminate  if  P,  Q,  and  R 
are  concurrent  or  parallel. 

In  PL  IV  is  worked  out  a  general  case,  and  the  deter- 
mination of  the  coordinates  of  the  required  center  of  moments 
is  seen  to  involve  considerable  labor.  Case  4,  in  its  occur- 
rence in  practice,  however,  almost  invariably  appears  with  the 


56  STATICS. 

three  unknowns  so  situated  that  the  coordinates  of  the  three 
centers  are  much  more  easily  determined  than  in  the  problem 
of  PL  IV.  The  arms  of  the  forces  can  usually  be  read  directly 
from  the  drawing  or  can  be  calculated  with  little  labor.  In 
such  cases  the  equations  will  best  be  written  with  the  arms 
inserted  directly  without  resort  to  components  except  for  the 
more  inconveniently  lying  forces. 

A  fairly  typical  example  of  Case  4  as  it  occurs  in  practice 
is  fully  worked  out  in  PL  V. 

Sometimes  two  of  the  unknowns,  say  P  and  <2»  are  paral- 
lel. R  can  then  be  found  by  taking  a  center  on  the  line  of 
action  of  Q  and  off  the  line  of  P  and  R,  bringing  into  the 
equation  the  previously  calculated  value  of  P. 

Graphic  Solution  (Fig.  n  and  cf.  PL  IV).  Let  the  knowns 
be  lettered  ab  .  .  .  .  cd,  and  the  unknowns  de,  ef,  and  fa, 
the  three  last  being  unknown  in  magnitude  (and  sense). 

Construct  the  magnitude  polygon  from  A  to  Dy  inclusive 
(Fig.  lib).  The  two  apices  E  and  /^remain  to  be  located. 
That  means  that  two  strings  oe  and  ^/"are  to  be  located,  while 
the  direction  of  neither  is  known,  nor  is  the  point  in  which  they 
will  intersect  ef  known.  One  of  these  directions  might  be 
assumed  and  the  string  polygon  closed  accordingly;  the  same 
result  could  be  attained  by  assuming  the  point  in  which  they 
intersect  ef.  The  result  of  this  process  would  in  general  be 
that  the  magnitude  polygon  would  not  close,  and  the  labor 
would  have  been  in  vain. 

Now,  if  either  of  these  two  strings  could  be  made  to  vanish, 
i.e.,  to  be  of  zero  length,  it  could  be  looked  upon  as  having 
any  direction  whatever,  including  a  direction  consistent  with 
the  closure  of  the  magnitude  polygon ;  then  only  one  string  re- 
maining to  be  located,  there  would  be  no  further  difficulty.  But 
any  string,  om,  common  to  two  forces,  Im  and  mn,  will  vanish 
if  ol  and  on  intersect  at  the  point  of  intersection  of  Im  and  mn. 


SOLUTION  OF  STATICAL  PROBLEMS. 


57 


Accordingly  if  oa  be  drawn  (Fig.   I  la)  at  the  outset  through 
the  intersection  of  ef  and  fa,  on  proceeding  with  the  string 


(6) 


FIG.  n. 

polygon  oe  will  fall  into  place,  connecting  the  point  where  od 
cuts  de  with  the  intersection  of  of  (whatever  its  direction)  and 
ef.  The  string  polygon  thus  being  closed,  E  is  located  at  the 
intersection  of  lines  from  D  and  O  parallel  respectively  to  de  and 
oe.  F  can  now  be  located  at  the  intersection  of  lines  from  E 
and  A  parallel  respectively  to  ef  and  af.  The  magnitude  poly- 
gon is  now  closed,  the  direction  of  the  ray  OF  is,  of  course, 
perfectly  consistent  with  the  string  polygon  being  closed,  and 
the  required  magnitudes  and  senses  are  DE,  EF,  and  FA. 


58  STATICS. 

Observe  that  the  key  to  this  solution  lies  in  drawing  the  first 
string  through  the  point  of  intersection  of  two  properly  selected 
consecutively  lettered  unknown  forces,  and  that  the  problem 
could  also  have  been  solved  by  starting  od  at  the  intersection 
of  de  and  efy  causing  oe  to  vanish,  F  to  be  located  by  aid  of 
the  string  of,  and  E  by  parallels  to  de  and  ef  from  D  and  F 
respectively. 

Observe  also  that  the  graphic,  unlike  the  algebraic,  solu- 
tion, is  not  essentially  more  laborious  in  an  instance  like  that 
of  PL  IV  than  in  PI.  V. 

5oa.  Remarks  on  Cases  3  and  4.  Case  3  may  be  looked  upon 
as  a  combination  of  Cases  4  and  I ,  for  any  two  convenient  com- 
ponents of  the  force  unknown  in  magnitude  and  direction  might 
be  substituted  for  this  force.  Then  they  and  the  other  un- 
known could  be  handled  by  Case  4,  and  the  two  components 
combined  into  the  single  required  force  by  Case  I.  In  fact  it 
not  uncommonly  happens  that  the  two  components  are  as  ac- 
ceptable a  result  in  practice  as  their  resultant  and  some  labor 
is  saved  by  substituting  them. 

In  a  similar  way  Case  4  may  be  looked  upon  as  a  combina- 
tion of  Cases  3  arid  2a.  The  intersection  of  the  two  unknowns 
may  be  regarded  as  the  given  point  of  application  of  the  re- 
sultant of  the  two  unknowns,  which  can  then  be  found  by  Case 
3,  and  resolved  into  its  components,  the  required  forces,  by 
Case  2a.  This  view  is  of  little  value  in  practice,  unless  per- 
haps in  throwing  additional  light  on  the  graphic  method  for 
solving  Case  4  given  in  the  preceding  section. 

Exercise  8.  Sketch  carefully  free  hand  the  graphic  solution  of  each 
of  the  four  cases,  assuming  the  necessary  data  for  each. 

Exercise  9.     Same  as  Exercise  8,  but  done  to  scale  as  usual. 

Exercise  10.  A  derrick  mast,  supported  by  the  usual  socket  and 
guys,  is  40  ft.  high  and  the  boom  is  60  ft.  long.  The  boom  is  held  at  an 
inclination  to  the  vertical  of  30  degrees  by  a  stay  running  from  its  upper 
end  to  the  top  of  the  mast.  If  a  weight  of  5000  Ibs.  be  suspended  from 


SOLUTION  OF  STATICAL  PROBLEMS. 


59 


the  end  of  the  boom,  what  would  be  the  forces  transmitted  through  the 
boom  and  stay  ? 

Solve  graphically  and  algebraically,  in  the  latter  case  using  the  Ax  and 
Ay  equations,  and  also  by  two  moment  equations  of  §  41. 

Exercise  n.  A  uniform  beam  rests  upon  supports  at  the  ends  at  the 
same  level.  The  length  of  the  beam  is  20  ft.  It  carries  vertical  loads  of 
10,  20,  30,  40  cwt.  at  5,  8,  13,  and  16  ft.  respectively  from  the  left  ends. 
Its  own  weight  is  5  cwt.  Determine  the  reactions  Ri  and  A'a  at  the  ends. 
Solve  by  both  methods. 

Suggestions.  Draw  the  data  diagram  to  the  scale  i  in.=  5  ft.  One 
data  diagram  serves  as  usual  for  both  methods  of  computation.  A 
single  heavy  line  will  suffice  for  representing  the  beam. 

Letter  forces  in  the  space  diagram  in  continuous  circuit  around  the  bar. 

For  the  algebraic  solution  apply  §  41,  determining  R±  and  R*  each  by 
an  independent  equation  in  the  form  given  in  PI.  \\b. 

Ri  and  R*  once  determined,  check  by  seeing  if  Ax  and  Ay  are 
satisfied. 

Exercise  12.  A  body  in  the  shape  of  an  isosceles  triangle  is  supported 
by  a  smooth  hinge  at  one  of  the  equal  angles  and  by  a  smooth  horizontal 
plane  (at  the  same  level  as  the  hinge)  at  the  other.  Assume  three  un- 
equal parallel  forces  normal  to  the  slope  on  the  side  next  the  hinge  to  be 
given,  and  determine  fully  the  pressures  on  the  hinge  and  plane.  Solve 
by  both  methods. 

Exercise  13.  Pi,  P*,  />»,  P4,  and  P*  (Fig.  12)  are  known  forces  acting 
on  the  body  shown,  and  Q,  R,  and  S  are  known  only  in  line  of  action.  If 


i 


FIG.  12. 

the  magnitudes  of  the  given  forces  are  4000,  3000,  2000,  1200,  and  I5oolbs. 
respectively,  determine  the  magnitudes  (and  senses)  of  Q,  R,  and  S  by 
both  methods. 

Suggestions.    Construct  the  data  diagram  carefully.    Scale  i  in.  =  5  ft. 


60  STATICS. 

See  PI.  IV  for  solution  by  the  usual  Axt  Ay,  and  £.  Observe  what  labor 
this  solution  involves. 

Use  the  method  of  §  41  and  solve,  scaling  arms  from  the  drawing 
only  for  checking.  In  writing  the  equation  it  will  be  convenient  to 
write  some  of  the  moments  in  the  form  Pxa,  and  some  in  the  form 
P  cos  a  y  —  P  sin  a.  x.  (Cf.  Plate  V.) 

Exercise  14.  A  square  plate  (Fig.  13)  weighing  900  Ibs.  is  held  in  a 
vertical  plane,  with  its  edges  inclined  30°  to  the  horizontal  and  vertical, 


ab 


FIG.  13. 

by  a  horizontal  force  and  forces  along  two  adjacent  sides  as  shown. 
Determine  these  three  forces  graphically  only. 

Suggestions.  Though  this  is  clearly  Case  4,  note  and  use  the  short 
cut  possible  in  this  special  case  where  there  are  two  pairs  of  forces  con- 
cerned whose  resultants  are  necessarily  equal,  opposite,  and  coincident. 
Putting  in  this  common  resultant — which  will  naturally  be  lettered  ac— 
there  are  two  groups  of  concurrent  forces,  ab,  be,  and  ac,  and  ac,  cd,  and 
da.  In  the  first  ab  is  known,  whence  be  and  ca  follow  by  Case  20  ;  ac 
can  then  be  resolved  similarly  into  £Y/and  da. 

Note  that  what  makes  this  exercise  a  special  case  is  that  there  is  only 
one  known  force.  The  method  just  pointed  out  maybe  regarded  as  the 
standard  method  of  resolving  a  force  graphically  into  three  components. 

The  algebraic  method  if  called  for  would  have  employed  the  three 
moments  equations  as  usual  in  Case  4. 

What  could  be  specified  as  to  the  position  of  the  pole  in  the  general 
method  of  Case  4  which  would  lead  to  a  solution  identical  with  this 
short  cut?  How  are  all  four  sides  of  the  string  polygon  then  ac- 
counted for? 

Exercise  15.  Resolve  (both  graphically  and  algebraically)  a  given 
force  of  any  convenient  assumed  magnitude  into  two  parallel  com- 
ponents whose  points  of  application  are 

(a)  On  opposite  sides  of  the  given  force. 

(b}  On  the  same  side  of  that  force. 


SOLUTION  OF  STATICAL  PROBLEMS.  61 

A 

Suggestions.  Avoid  a  small  scale  for  the  space  diagram.  As  is  usual 
in  this  Case,  the  equations  of  moments  are  to  be  preferred  to  the 
Ax,  Ay,  and  B  equations  for  the  algebraic  work. 

NOTE. — For  further  general  problems  in  statics,  the  reader  is  referred  to  such 
works  as  Loney's  Statics,  Bowser's  Analytical  Mechanics,  Minchin's  Statics,  Wal- 
ton's Problems,  etc.  The  reader  seeking  practice  among  the  problems  there  given 
must  be  prepared  in  many  of  those  problems  for  a  large  amount  of  geometrical 
analysis  and  trigonometric  reduction  extraneous  to  the  purely  statical  solution. 


CHAPTER   VII. 
ADDITIONAL  GENERAL  TOPICS  AND  PROCESSES. 

51.  Graphic  Representation  of  the  Moment  of  a  Force. 
— THEOREM.  If,  through  any  point,  a  line  be  drawn  parallel 
to  a  given  force,  P,  the  distance,  jj/,  intercepted  from  this  line 
by  the  two  strings  belonging  to  the  force  is  a  length  such  that 
when  multiplied  by  the  force  H,  measured  by  the  perpendic- 
ular dropped  from  O  upon  the  given  force  in  the  magnitude 
diagram,  the  result  will  be  numerically  equivalent  to  the  mo- 
ment of  P  about  the  given  point. 

Proof. — -Let  m  (Fig.    14)  be  any  point  whose  perpendic- 


y 


FIG.   14. 

ular  distance  from  the  line  of  action  of  P  is  /.  If  P  be  lettered 
ab,  and  the  pole  be  taken  and  the  string  polygon  be  con- 
structed as  shown,  y  is  a  length  such  that 

Hy  =  Pp. 

62 


ADDITIONAL   GENERAL    TOPICS  AND  PROCESSES.  63 

• 
For,  noting  the  similarity,  by  construction,  of  the  triangles 

OAB  and  the  one  bounded  by  oa,  ob,  and  the  line  through  m, 
it  appears  that 

AB  :  H  =  y  :  py 

or  as  AB  represents  the  magnitude  of/5, 

Hy  =  Pp.  Q.  E.  D. 

The  intercept  y  may  accordingly  be  said  to  represent  the 
moment  of  P  about  m,  it  being  understood  that,  to  get  the 
numerical  value  of  this  moment,  y  measured  in  the  proper 
scale  of  lengths  must  be  multiplied  by  H  measured  in  the 
proper  scale  of  force  magnitudes. 

So  far  H  and  y  are  simply  a  force  and  a  distance  whose 
product  is  the  same  as  that  of  P  and/.  There  are  of  course 
an  indefinite  number  of  forces  and  distances  whose  products 
will  have  this  value,  and  the  substitution  if  desired  might  be 
made  in  simpler  ways  even  than  by  this  theorem,  if  that  were 
all  that  is  desired.  With  a  series  of  parallel  forces,  however, 
treated  as  usual  with  the  magnitude  and  string  polygons,  H 
will  be  constant  for  them  all,  and  aided  by  this  fact  the  theorem 
leads  to  a  convenient  graphical  method  for  the  treatment  of 
moments  which  will  be  developed  in  the  next  section. 

Observe  that  the  strings  oa  and  ob  are  simply  any  two  com- 
ponents of  P,  and  that  H  is  merely  their  common  component 
normal  to  P. 

52.  String  Polygon  for  Parallel  Forces  a  Diagram  of 
Moments. — As  a  corollary  to  the  theorem  of  the  preceding 
section,  it  may  be  stated  that  with  parallel  forces  any  two 
sides  of  the  string  polygon,  extended  if  necessary,  will  cut  from 
a  line  parallel  to  the  forces  an  intercept  which  will  represent 
the  sum  of  the  moments  (about  any  point  in  that  line)  of  all 
the  forces  at  the  apices  of  the  polygon  included  between  the 
two  sides. 


04  STATICS. 

For  example,  suppose,  in  Fig.  15,  the  string  polygon  be 
drawn  for  any  set  of  parallel  forces  (magnitude  polygon  not 
shown)  which  for  the  sake  of  generality  are  not  taken  in  equi- 
librium and  which  are  lettered  preferably,  though  not  necessa- 
rily, in  the  order  of  their  occurrence.  Lengths  are  intercepted 
on  the  line  MN,  parallel  to  the  forces,  which  represent  the 


moments  of  groups  of  these  forces  about  any  point  m  on  MN 
as  follows. 

The  intercept  a^a^  bounded  by  oa  and  og  represents  the 
moment  of  ag  by  direct  application  of  §  51  ;  that  is,  it  repre- 
sents the  moment  of  the  resultant  of  the  whole  set,  and  hence 
the  sum  of  the  moments  of  all  the  forces  of  the  set. 

Similarly,  a^a3  represents  the  moment  of  ad\  that  is,  the  sum 
of  the  moments  of  ab,  be,  and  cd.  Likewise  a2a3  represents 
the  moment  of  dg\  that  is,  of  de,  ef,  andy^  combined. 

Moreover,  the  intercepts  are  the  actual  summations  of  other 
intercepts  each  representing  the  moment  of  one  of  the  indi- 
vidual given  forces.  Take,  for  instance,  the  last  case,  that  of 
de,  efy  andy%\  Extending  the  strings  oe  and  of  till  they  cut 
MN,  the  resulting  intercepts  aza± ,  a4a5 ,  and  a5a2  are  seen  by 


ADDITIONAL   GENERAL    TOPICS  AND  PROCESSES.  65 

0 

§  5 1  to  represent  the  moments  of  de,  ef,  and  fg  respectively 
about  m,  and  what  is  more,  a^a^  is  their  algebraic  sum  (—  #3tf4+ 
#5tf4— <25#2),  as  it  should  be. 

The  force  //,  it  may  be  repeated,  by  which  the  lengths  of 
these  various  intercepts  must  be  multiplied  to  give  the  numeri- 
cal values  of  the  respective  moments,  is  the  force  represented 
by  the  perpendicular  (measured  in  the  same  scale  as  the  rest 
of  the  forces)  dropped  from  0  upon  the  straight  line  A  BCD, 
etc.,  upon  which  fall  the  parallel  forces  ab>  be,  cd,  etc.,  when 
they  appear  in  the  magnitude  polygon,  //"being  thus  a  con- 
stant force,  the  moments  are  proportional  to  the  intercepts  as 
above  stated,  and  the  string  polygon  for  the  parallel  forces  is  a 
diagram  of  moments. 

53.  Remarks  on  the  String  Polygon  as  a  Diagram  of 
Moments. — Referring  still  to  Fig.  15,  it  will  be  useful  to  note 
the  following  facts  and  deductions. 

If  the  forces  had  been  in  equilibrium,  og  and  oa,  the  strings 
including  the  whole  set,  would  have  coincided  and  there  would 
have  been  no  difference  between  a^B  and  a2ay  except  in  sign, 
and  their  sum  would  be  found  to  be  zero,  as  it  should  be. 

If  the  forces  had  been  reducible  to  a  couple,  oa  and  og 
would  have  been  parallel  and  a^a2  constant,  as  it  should  be,  for 
the  constant  moment  of  a  couple. 

Where  oa  and  og  intersect,  a^2  would  vanish,  as  it  should 
do,  for  that  intersection  is  known  to  be  on  the  line  of  action  of 
the  resultant  of  the  set  of  forces. 

If  the  forces  in  the  given  set  are  not  parallel,  the  string  poly- 
gon can  still  be  made  to  yield  the  sum  of  the  moments  about 
a  point  of  any  consecutively  lettered  group  of  these  forces.  It 
is  only  necessary  to  draw  a  line  through  the  point  parallel  to 
their  resultant,  scale  the  intercept  from  this  line  by  the  strings 
inclosing  the  group,  and  multiply  it  by  the  //"for  this  particu- 
lar resultant.  The  process  is  a  straightforward  application  of 


66  STATICS. 

§  5 1.  Since  in  general  no  two  resultants  will  have  the  same  ff, 
or  be  in  the  same  direction,  the  string  polygon  for  non-parallel 
forces  is  not  a  diagram  of  moments  in  the  same  useful  sense  as 
\vhen  the  forces  are  parallel. 

Hence  the  string  polygon  is  not  only  an  equivalent  to  an 
equation  of  moments  in  its  scope  and  uses,  but  also  an  exact 
parallel  to  it  in  that  it  readily  gives  the  value  of  any  term  of 
that  equation  and  is  simply  a  graphical  means  of  finding  and 
summing  the  values  of  those  terms. 

If  any  set  of  forces  whose  magnitude  polygon  closes  is  not 
reducible  to  a  couple,  the  intercept  from  any  line  whatever  in 
the  plane  by  the  two  extreme  strings  must  be  zero.  This  could 
not  be  unless  these  two  strings  coincide,  or,  in  other  words, 
unless  the  string  polygon  close,  a  fact  the  converse  of  which 
could  be  taken  as  a  demonstration  of  the  preclusion  of  rotation 
by  the  closure  of  a  string  polygon  alternative  to  that  of  §  32. 

If  the  set  of  forces  is  reducible  to  a  couple,  the  resultant 
looked  upon  as  a  single  force  is  indeterminate  in  direction  (cf. 
Exercise  7).  Nevertheless  its  moment  can  still  be  found  from 
the  string  polygon  by  assigning  it  any  convenient  direction, 
taking  the  intercept  by  the  extreme  strings  (lettered,  say,  oa  and 
on)  from  a  parallel  to  it  and  multiplying  this  intercept  by  the 
H  measured  by  the  perpendicular  from  0  dropped  upon  a  par- 
allel to  the  intercept  drawn  through  the  two  coinciding  points 
A  and  TV  in  the  magnitude  polygon.  The  process  by  which 
the  result  was  evaluated  in  Exercise  7  may  be  regarded  as 
really  an  application  of  this  method,  in  which  a  direction  normal 
to  the  extreme  strings  was  assigned  to  the  resultant,  making 
//"identical  with  OA. 

54.  To  Pass  a  String  Polygon  Through  Three  Given 
Points. — A  process  called  passing  a  string  polygon  through 
three  given  points  is  sometimes  useful  in  the  study  of  the  equi- 
librium of  structures.  It  consists  in  locating  a  pole  for  a  set 


ADDITIONAL   GENERAL    TOPICS  AND  PROCESSES. 


67 


of  two  or  more  forces  so  that  three  specified  strings  of  the  result- 
ing string  polygon  will    each    (prolonged    if  necessary)  pass 
through  one  of  three  given  points  not  in  the  same  straight  line.* 
Problem. — Suppose,  for  example,  that  the  forces  in  Fig.  16, 


ab,  be,  cdj  and  de,  their  magnitude  polygon  ABCDE,  and  the 
three  points  mv  mv  and  m^  be  all  given,  and  that  it  is  required 
to  select  a  pole  O  so  that  oa,  oc,  and  oe  will  pass  respectively 
through  mv  mv  and  my 

Solution.- — Taking  a  pole   0'  at  random,  determine  (Case 
3)  two  forces   CCla.nd  ACV  the  former  acting  at  m2  in  any 

*If  there  is  only  one  force  in  the  given  set,  the  string  polygon  can  still  be 
made  to  pass  through  three  points  not  in  a  straight  line,  but  two  of  the  points  will 
of  course  have  to  be  on  one  of  the  strings. 


68  STATICS. 

convenient  direction,  and  the  latter  at  mt  in  a  direction  deter- 
mined accordingly,  such  that  they  and  the  given  forces  ab  and 
be,  intervening  between  ml  and  mz,  will  constitute  a  set  in 
equilibrium*  Now,  when  oc  passes  through  m2  as  required,  ocl 
will  have  to  pass  through  mz  and  m^  in  order  to  close  the  string 
polygon  for  this  balanced  set.  It  follows,  moreover,  for  the 
same  reason,  that  if  ocl  passes  through  m^  and  mv  oc  will  have 
to  pass  through  mz,  and  working  backwards,  oa  will  have  to 
intersect  ocl  again  at  mr  Therefore  an  O  taken  anywhere 
on  a  line  through  Cl  parallel  to  mlm2  will  be  one  for  which,  if 
oa  be  drawn  through  mv  oc  will  pass  through  m¥ 

Repeat  this  process,  determining  an  AEl  and  an  EEl  at  ml 
and  ^respectively,  which  would  balance  ab,  be,  cd,  and  de, — 
AE^  and  EEl  corresponding  exactly  to  the  ACl  and  CC^ 
respectively  of  the  preceding  step.  Then  reasoning  as  before, 
an  O  taken  anywhere  on  a  line  through  El  parallel  to  m^n^  will 
be  one  for  which,  if  oa  be  drawn  through  mr  oe  will  pass 
through  my 

Therefore  the  intersection,  O,  of  lines  from  Cl  and  El  par- 
allel respectively  to  m1m2  and  m1mB  will  be  the  required  pole 
for  which  if  oa  be  passed  through  mv  oc  will  pass  through  mz 
and  oe  through  mz  as  required.  The  string  polygon  oa,  ob,  oc, 
od,  oe  can  now  be  constructed  at  leisure. 

The  foregoing  process  may  be  described  in  general  terms 
as  follows. 

If  for  a  set  of  given  forces,  ab,  be,  cd  .  .  .  yz,  it  be  required 
to  pass  any  three  specified  strings,  od,  ol,  and  os,  of  their  string 
polygon  respectively  through  the  given  points  mv  mz,  and  mB 
not  in  one  straight  line,  draw  a  string  polygon  at  random 
from  a  pole  O'  selected  at  random,  beginning  with  the  string 
o'd,  making  that  pass  through  mv  and  by  working  both  ways 
complete  this  random  polygon  at  least  as  far  as  o's.  Use  this 
string  polygon  to  determine  (Case  3)  //j  and  dlv  and  ssl  and 


ADDITIONAL   GENERAL   TOPICS  AND  PROCESSES.  69 

• 

dsl  (//!  and  ss^  being  assumed  to  act  at  m2  and  mz  respectively 
in  any  convenient  direction,  and  the  other  two  forces  at  m^ 
and  determined  accordingly),  which  would  balance  de  .  .  .  /£/, 
and  de  .  .  .  rs  respectively.  DL^  and  DS^  do  not  need  to  be 
drawn.  Parallels  to  mjn^  and  mjn^  through  L^  and  Sl  re- 
spectively will  locate  the  required  pole  0  at  their  intersection. 
To  draw  the  required  string  polygon,  od  will  be  drawn  first  and 
made  to  pass  through  mr  Then  working  both  ways,  the 
polygon  is  completed. 

The  problem  obviously  does  not  admit  of  solution  when  the 
three  points  are  in  a  straight  line. 

It  is  advisable  to  take  0'  so  that  the  random  string  poly- 
gon will  lie  as  well  as  possible  out  of  the  way  of  the  final 
required  string  polygon. 

The  random  string  polygon  can  be  obviated  if  desired  by 
an  algebraic  computation  of  the  magnitude  and  direction  of  such 
a  force  as  OD  of  the  preceding  general  problem.  D  being 
given,  0  can  then  be  plotted  directly. 

To  determine  OD  we  need  to  observe  that  its  point  of  appli- 
cation ml  is  given,  and  that  the  resultant  of  it  and  de  .  .  .  /£/, 
and  of  it  and  de  .  .  .  rs  must  pass  through  m2  and  m^  respec- 
tively. 

It  will  be  well  to  consider  od  replaced  by  its  horizontal  and 
vertical  components  meeting  at  mv  whose  magnitudes  and 
senses  *  can  be  determined  from  two  simultaneous  equations  of 
moments  for  centers  respectively  at  m2  and  my  the  first  one  for 
the  series  offerees  de  .  .  .  /£/and  the  unknowns,  the  latter  for 
the  series  de  .  .  .  r s  and  the  unknowns. 

Combining  the  two  components  (by  Case  20),  the  required 
OD  is  found,  and  O  can  be  plotted. 

*It  will  be  found  advisable,  in  order  to  avoid  error  in  such  simultaneous 
equations,  to  assume  senses  of  the  components  arbitrarily  and  show  them  in  the 
sketch  of  data  subject  to  subsequent  correction  on  the  solution  of  the  equation. 


70  STATICS. 

It  will  be  highly  desirable  in  practice  to  check  this  loca- 
tion of  O  by  determining  in  a  similar  manner  the  magnitude 
and  direction  of  5(9.  The  centers  will  be  at  m2  and  ml  and  in- 
cluded in  the  equations,  besides  the  two  unknown  components 
will  be  the  sets  sr  .  .  .  ml,  and  sr  .  .  .  ed  respectively. 

The  object  of  the  whole  process  is  sometimes  merely  the 
determination  of  forces  at  two  of  the  points,  such  as  the  OD  and 
OS  of  the  preceding  work.  The  foregoing  section  constitutes 
therefore  a  complete  description  of  the  solution  of  such  prob- 
lems by  both  the  algebraic  and  graphic  methods.  (Cf.  §  86.) 

Exercise  16.  Assume  any  set  of  seven  vertical  forces  uniform  in  sense 
and  letter  them  in  the  order  of  their  occurrence  ab,  .  .  .  hi.  Assume 
three  points  m\,  m9 ,  and  ms  not  in  the  same  straight  line  and  at  the  left 
of  ad,  between  de  and  ef,  and  at  the  right  of  hi  respectively.  Pass  a 
string  polygon  for  the  assumed  forces  through  the  three  assumed  points 
so  that  oa,  oe,  and  oi  will  go  through  m\ ,  m* ,  and  m*  respectively. 
Solve  graphically  only. 

Remark.  Assuming  the  set  of  forces  as  specified  does  not  interfere 
with  the  generality  of  the  method  required  and  will  save  some  labor. 

55.  An  Alternative  View  of  the  Rays  and  the  String 
Polygon. — Each  two  successive  rays  may  be  regarded  as  two 
components  of  the  force  with  which  they  form  a  closed  force 
triangle,  and,  moreover,  by  reversing  them  in  sense,  as  com- 
ponents respectively  of  the  forces  next  preceding  and  follow- 
ing that  force  in  the  magnitude  polygon.  Inserting  these 
pairs  of  components  in  place  of  their  respective  resultants,  there 
will  result  a  set  equivalent  to  the  given  set.  If  by  moving  the 
assumed  point  of  intersection  of  the  components  along  the  lines 
of  action  of  their  resultants  the  pairs  of  equal  and  opposite  com- 
ponents of  the  consecutively  lettered  forces  can  all  be  made 
coincident,  the  components  must  be  a  set  in  equilibrium,  and 
hence  the  original  set  must  have  been  as  well. 

The  fact  that  the  components  were  pairs  of  equals  as  well 
as  opposites  could  only  be  if  the  magnitude  polygon  closed. 


ADDITIONAL   GENERAL    TOPICS  st*u  r  ACCESSES.  71 

• 

The  pairs  all  coinciding  form  the  closed  string  polygon.  The 
lines  of  action  of  pairs  of  coinciding  forces  constitute  the 
strings. 

If  the  pairs  of  equal  and  opposite  components  all  coincide 
and  neutralize  one  another  but  one,  that  pair  of  components 
constitutes  a  couple  to  which  the  given  set  is  reducible. 

If  the  set  is  not  reducible  to  a  complete  set  of  pairs  of  equal 
and  opposite  components  (a  case  which  can  only  arise  when  the 
magnitude  polygon  fails  to  close),  there  will  be  only  one  pair 
not  of  this  character.  The  pairs  of  equals  and  opposites  could 
be  made  to  coincide  throughout  in  the  space  diagram  as  usual, 
leaving  the  other  two  to  locate  their  resultant  passing  through 
their  intersection,  a  resultant  likewise  that  of  the  given  system. 

From  the  preceding  three  paragraphs  could  be  dedyced 
anew  the  principle  that  the  closure  of  the  magnitude  polygon 
and  the  string  polygon  insures  equilibrium. 


PART  II. 

APPLICATIONS. 


CHAPTER   VIII. 
CENTERS   OF   GRAVITY. 

.56.  Center  of  Gravity.— The  center  of  gravity  of  a  body 
is  that  point  through  which  the  resultant  of  the  gravity  forces 
acting  upon  all  parts  of  the  body  will  pass,  whatever  the  atti- 
tude of  the  body,  or  what  amounts  to  the  same  thing,  whatever 
we  may  regard  the  slope  of  the  gravity  forces,  so  long  as  they 
remain  parallel.*  The  center  of  gravity  may  be  looked  upon, 
then,  as  the  invariable  point  of  application  of  the  force  which 
represents  the  weight  of  the  body. 

Since  the  moment  of  the  weight  of  the  body,  referred  to  any 
plane,  is  evidently  equal  to  that  of  the  weights  of  all  the  parts 

*  Strictly  speaking,  the  center  of  gravity  is  a  point  in  which  the  whole  mass  of 
a  body  might  be  concentrated  without  affecting  gravity  attractions  existing  between 
the  body  and  all  other  bodies,  whatever  may  be  their  relative  distance  and 
position. 

Very  few  bodies  have  a  true  center  of  gravity.  What  is  commonly  called  the 
center  of  gravity  would  more  properly  be  called  center  of  mass  or  center  of  area. 

All  centers  of  gravity  are  also  centers  of  mass,  but  all  centers  of  mass  are  not, 
strictly  speaking,  centers  of  gravity. 

The  term  center  of  gravity  is  so  thoroughly  fixed  by  general  usage  in  the  sense 
above  given,  and  the  need  for  making  the  distinction  is  so  rare,  that  there  is  no 
great  need  for  urging  the  substitution  of  the  more  accurate  terms.  For  further 
discussion  of  this  distinction  see  Du  Bois,  Mechanics,  vol.  2,  chap.  4. 

72 


CENTERS  OF  GRAVITY.  73 

referred  to  the  plane,  we  can  write  (conceiving  gravity  acting 
parallel  to  the  plane)  : 


that  is 

likewise  y2w  —  ^(wy)      -  ,     .      .      .      .      (i') 

also 


where  !r,  J/,  ~z  and  x,  y,  z  are  the  coordinates  of  the  centers  of 
gravity  of  the  whole  and  of  parts  of  the  body  respectively,  and 
w  the  weights  of  these  parts. 

It  is  often  required  to  find  ~x,  ~y,  z,  and  to  do  it  we  need 
only  to  divide  the  body  into  parts  whose  x,  y,  z  can  readily  be 
determined,  and  apply  from  (i') 


_ 


Equations  (i)  show  that  for  all  planes  or  axes  of  reference 
passing  through  this  center  of  gravity,  i.e.  for  ~x  =  o,  ~y  =  o, 
and  #"=  o, 

2(wx}  =  o;     2(wy)  =  o;     2(wz)  =  o;   .      .      .      (2) 

as  might  have  been  foreseen. 

It  is  often  convenient  to  apply  the  term  center  of  gravity 
to  bodies  irrespective  of  their  weights  or  to  those  having  no 
weights,  as  geometrical  shapes,  and  planes  and  lines,  meaning 
thereby  the  point  which  would  be  the  center  of  gravity  of  the 
body  if  the  body  were  of  uniform  density  or  had  weight  pro- 
portional to  volume,  area,  or  length.  Any  factor  introduced 
to  express  weights  in  terms  of  volumes,  areas,  and  lengths 
would  cancel  out  of  equations  (i)  and  (2)  and  leave  simply 
volumes  and  areas  and  lengths  in  place  of  weights,  to  be  treated 
exactly  like  weights.  In  (i)  and  (2),  then,  might  be  substi- 
tuted Vj  a,  or  /  for  w.  In  the  following  work  attention  will  be 
devoted  mainly  to  the  center  of  gravity  of  areas,  and  they  like 


74  STATICS. 

lines  will  preferably  be  referred  to  two  rectangular  axes,  and  2 
will  disappear. 

In  the  foregoing  the  term  "  system  of  bodies  "  (if  the  bodies 
be  understood  to  be  fixed  relatively  to  one  another)  might  be 
substituted  for  * '  body. ' ' 

Statically  the  location  of  a  center  of  gravity  is  the  location 
of  the  point  of  application  of  the  resultant  of  a  set  of  parallel 
forces,  a  simple  special  case  under  Case  I  requiring  only  the 
familiar  equations  (of  which  (i)  above  is,  of  course,  a  deriva- 
tive) and  polygons. 

In  the  case  of  simple  and  symmetrical  bodies  the  result  of 
applying  equations  or  polygons  can,  of  course,  be  seen  at  once 
by  mere  inspection,  just  as  can  sometimes  be  done  with  other 
statical  problems. 

The  center  of  gravity  of  a  straight  line  is  evidently  its 
middle  point;  of  a  rectangle  is  the  intersection  of  normals  to 
the  centers  of  its  sides ;  of  any  parallelogram  the  intersection 
of  its  diagonals ;  of  a  circle  or  ellipse  its  geometrical  center ;  of 
a  triangle  the  intersection  of  its  medial  lines.  This  opens  the 
way  for  proof  by  the  general  method  that  the  center  of  gravity 
of  a  trapezoid  is  the  intersection  of  the  line  connecting  the 
middle  points  of  its  parallel  sides  with  the  diagonal  connecting 
points  found  by  extending  each  of  the  parallel  sides  in  opposite 
directions  by  an  amount  equal  to  the  length  of  the  opposite 
side,  or  at  the  intersection  of  two  such  diagonals.  (Fig.  17.) 

-——a——a--—-      21) * 


___6._._^ 

FIG.  17. 

The  center  of  gravity  of  a  surface  bounded  by  a  curved  line 
is,  of  course,  treated  by  equation  (i),  but  for  strict  solutions 


CENTERS  OF  GRAVITY.  75 

jf 
in  such  cases  subdivisions  w  are  made  infinitesimally  small  and 

the  problem  falls  into  the  field  of  the  integral  calculus  rather 
than  arithmetic,  without,  however,  introducing  any  new  stati- 
cal principle.  The  solution  may  be  approximated  with  any 
desired  degree  of  exactness  by  arithmetical  methods  by  mak- 
ing the  parts  w  smaller  and  smaller. 

In  the  case  of  a  segment  of  a  parabola  cut  off  by  a  chord 
normal  to  the  axis,  if  b  be  the  length  of  such  chord  and  h  its 
distance  from  the  vertex,  integration  will  show  that  its  area  is. 
\bh,  and  that  its  center  of  gravity  is  on  the  axis  distant  \h 
from  the  vertex  (Fig.  18).  The  center  of  gravity  of  a  circular 

I 


FIG.  18. 
sector  is  shown  by  similar  means  to  be  distant  from  the  center 

of  the  circle  on  the  central  radius  by  \r  — -5—,  where  r  is  the  *?- 

radius  of  the  circle  and  0  is  half  the  central  angle  of  the  sec- 
tor in  circular  measure. 

Any  plane  surface  which  can  be  divided  with  sufficient' 
exactness  into  triangles,  rectangles,  trapezoids,  parallelo- 
grams, segments  of  parabolas,  or  sectors  of  circles  can  accord- 
ingly be  treated  by  the  general  method.  For  a  surface  it  is 
necessary  in  general  to  assume  the  gravity  forces  acting  first 
in  one  direction,  then  in  another,  it  may  well  be,  at  right  angles 
to  the  first.  The  point  common  to  the  two  resultants  for  the 
two  different  directions  for  the  forces  will  be  the  center  of 
gravity. 

Numerical  example.  Required  the  location  of  the  center 
of  gravity  of  the  irregular  surface  shown  in  Fig.  1 9. 

Solution.  Taking  the  origin  at  the  lower  left-hand  corner, 
and  dividing  the  figure  into  two  rectangles  and  a  semicircle,  the 
areas  and  the  coordinates  of  the  centers  of  gravity  of  each  of 


76 


STATICS. 


these  divisions  are  worked  out  and  recorded  in  the  figure  for 
convenience  in  reference.     The  center  of  gravity  of  the  semi- 


"h/ 


fr 10i' 


FIG.  19. 

circle  is,  according  to  what  has  been  said  just  above  as  to  the 
center  of  gravity  of  a  sector, 


2  X  4  X  sin  - 

7t 
3*2 


8 


4.71 


=  1.70  in. 


above  the  center  of  the  circle,   making  the  ordinate  of   the 

center  of  gravity  2.0  +  4-O  +  1.70=  7.70  in.  as  shown. 

Algebraic  method.    Calling  ;r  and  y  the  coordinates  of  the 

center  of  gravity  of  the  whole  figure,  the  algebraic  solution  can 

conveniently  be  arranged  as  follows : 

25.1  X  4-0  =  100.40         25.1  X  7-7  =  193-27 
32.0  x  4-0  —  128.00         32.0  X  4.0  =  128.00 

20.0  X   5-0  =   100.00  20.0  X    I.O  =      20.00 


77.i  X  x     =  328.40 

328.40  .  _       341-27 

x  =  -    =  4-2O  in.  y  = =  4.43  in. 

77.1  '          77.1         ^^ 


CENTERS  OF  GRAVITY.  77 

0~ 

Graphic  method.  Considering  forces  proportional  to  the 
areas  of  the  divisions  acting  in  two  directions  at  right  angles  to 
each  other  from  the  centers  of  gravity  of  their  respective  divi- 
sions, and  locating  the  resultant  of  each  of  these  sets  of  parallel 
forces  by  the  usual  method,  the  required  center  of  gravity  is 
found  at  the  intersection  of  the  two  resultants.  This  work  is 
done  in  Fig.  19,  with  results  agreeing  with  those  found  above. 
One  magnitude  polygon  only  is  required.  By  transferring  nor- 
mals to  its  rays  as  well  as  parallels,  both  string  polygons  can 
be  constructed. 

The  reader  interested  in  a  wider  range  of  applications  of 
the  principles  of  this  chapter  is  referred  to  works  such  as  those 
mentioned  at  the  close  of  Chapter  VI,  in  addition  to  which 
might  well  be  mentioned  Rankine's  Applied  Mechanics,  or 
Weisbach's  Mechanics. 

Exercise  17.  Find  the  center  of  gravity  of  any  irregular  figure 
bounded  by  curves  and  straight  lines.  Solve  both  graphically  and  alge- 
braically. 

Suggestion.  Select  the  figure  from  among  the  rolled-steel  sections, 
such  as  the  channels,  unequal-legged  angles,  bulb  angles,  deck-beams, 
etc.,  given  in  the  Carnegie,  Cambria,  Pencoyd,  or  other  manufacturers' 
handbooks.  Results  may  then  be  compared  with  those  given  in  the 
handbook. 


CHAPTER  IX. 
STRESS. 

57.  External  and  Internal  Forces. — External  forces  upon 
a   body    are    forces    exerted    upon  that  body  by  or   through 
another  body. 

Internal  forces  in  a  body  are  those  transmitted  to  one  part 
of  that  body  through  another  part  of  the  same  body. 

Investigation  of  external  forces  determines  whether  a  body 
as  a  whole  is  stable,  without  regard  to  whether  all  parts  are 
strong  enough  to  do  what  is  required  of  them.  Investigation 
of  internal  forces  determines  the  strength  required  of  the  parts 
to  insure  the  stability  of  the  whole.  A  knowledge  of  both 
external  and  internal  forces  is  therefore  indispensable.  Both 
internal  and  external  forces  are,  of  course,  subject  to  the  gen- 
eral laws  governing  forces. 

The  internal  forces  at  any  section  of  the  body  hold  in  equi- 
librium the  external  forces  on  either  side  of  section. 

58.  Stress. — Stress  is  the  tendency  to  distortion  or  rupture 
in  a  body  due  to  the  action  of  the  external   forces.      It  is  a 
measure  of  the  responsibility  of  the  body  or  of  a  specified  part 
of  the  body  in  service. 

Stress  may  vary  greatly  in  the  various  parts  of  the  body, 
and  it  is  necessary  in  any  study  of  it  to  compute  the  stresses 
in  existence  at  one  or  more  plane  sections  of  the  body.  In 
such  cases  the  plane  of  the  section  will  divide  the  body  into 
two  segments,  and  the  external  forces  into  two  sets,  one  acting 

78 


STRESS.  79 

0 

directly  on  one  segment,  the  other  on  the  other.  The  two 
equal,  opposite,  and  coincident  resultants  of  these  two  sets  of 
forces  may  be  looked  upon  as  the  immediate  cause  of  the  stress 
and  its  measure  as  well.  The  magnitude  of  the  stress  depends 
upon  the  common  magnitude  of  these  two  resultants,  and  its 
nature  upon  their  other  elements. 

The  determination  of  stresses,  then,  is  simply  the  deter- 
mination of  elements  of  forces — a  statical  process  pure  and 
simple  and  one  of  the  most  important  fields  of  application  of 
statics.  It  will  receive  much  attention  accordingly. 

59.  Kinds  of  Stress. — The   resultants   mentioned  in  the 
preceding  section  may,  of  course,  be  either  single  forces  or 
couples,  and  four  special  cases  of  stress  are  distinguished,  each 
corresponding  to  a  special  direction  for  these  single  forces  and 
to  a  special  position  for  the  planes  of  the  couples. 

With  the  single  force  resultants  there  is  Normal  Stress  or 
Tangential  Stress  (more  commonly  called  Shear)  at  the  sec- 
tion according  as  the  resultants  are  normal  or  parallel  to  the 
section.  Their  common  magnitude  measures  each  stress. 

With  the  couples  there  is  Flexure  or  Torsion  at  the  sec- 
tion according  as  the  couples  are  in  a  plane  normal  to  the  sec- 
tion or  in  planes  parallel  to  it.  The  common  magnitude  of 
the  couples  measures  the  stress. 

60.  Combined   Stresses. — As  a  matter  of  fact,  however, 
the  resultants  are  very  frequently  not  in  any  one  of  these  four 
simple  relations  to  the  section,  and  there  results  some  combi- 
nation of  the  four  stresses  accordingly.      If  the  single  force  re- 
sultants are  inclined  to  the  section  there  exists  both  normal 
stress  and  tangential  stress,  each  measured  by  the  common 
magnitudes  of  the  components  normal  and  parallel  to  the  sec- 
tion respectively.      If  such  resultants  fail  to  pass  through  the 
center  of  gravity  of  the  section  it  will  be  convenient  to  observe 
that  by  §  20  each  resultant  is  equivalent  to  a  force  acting  at 


8o  STATICS. 

that  point*  and  a  couple,  and  in  such  cases  accordingly  there 
would  be  combined  normal  stress,  shear,  and  flexure.  If 
the  forces  are  non-coplanar,  there  may  be  still  another 
couple  in  a  plane  parallel  to  the  section,  and  torsion  will  be 
present  in  addition  to  the  other  stresses. 

If  there  is  normal  stress  or  shear  at  the  section  as  well  as 
flexure,  as  is  usually  the  case,  the  resultant  couple  which  will 
measure  the  flexure  is  the  resultant  of  the  set  made  up  of  all  the 
forces  external  to  the  body  acting  on  the  segment  and  also  a 
force,  internal  to  the  body  but  external  to  the  segment,  equal, 
opposite,  and  parallel  to  these  external  forces  conceived  to  be 
applied  at  the  center  of  gravity  of  the  section. t 

In  other  words,  flexure  may  be  looked  upon  as  measured 
by  the  couple  remaining  after  providing  the  forces  which  will 
prevent  motion  of  the  segment  normal  or  parallel  to  the  section. 

61.  Further  Particulars  Relating  to  Stress Normal 

stress  is  divided  into  its  more  familiar  subdivisions,  compres- 
sion and  tension,  according  as  the  resultants  act  from  their 
respective  segments  towards  or  away  from  the  section. 

Shear  and  Flexure  are  similarly  subdivided  according  to  the 
relative  senses  of  the  resultants,  but  their  subdivisions  have  no 
established  names,  and  are  distinguished  from  each  other  by 
the  adjectives  positive  and  negative,  arbitrarily  applied.  These 
distinctions  are  rarely  needed  except  in  the  case  of  horizon- 
tal beams  subject  to  vertical  loads,  in  which  the  sections 

*  Any  other  convenient  point  might  be  selected,  but,  as  a  principle  of  Resistance 
of  Materials  worth  noting  in  passing,  it  may  be  said  that  there  is  no  advantage  in 
making  the  substitution  of  the  force  and  the  couple  unless  the  center  of  gravity  be 
the  point  of  application  of  the  force.  It  is  known  how  to  provide  for  a  force  so 
applied,  but  not  for  one  elsewhere,  except  by  this  substitution. 

j-  If  this  force  be  taken  into  account  the  value  of  the  flexure  would  be  independ- 
ent of  the  center  chosen,  as  the  value  of  a  couple  is  constant  for  all  centers  in 
the  plane  (§  23).  But  taking  the  center  of  moments  at  the  center  of  gravity  the 
moment  of  this  force  vanishes  and  the  force  need  not  then  be  determined  for  the 
calculation  of  the  flexure. 


STRESS. 


Si 


are  taken  normal  to  the  axis  of  the  beam,  i.e.  vertical.  In 
such  cases  it  is  natural  and  customary  to  call  shear  and  flexure 
positive  when  the  left-hand  segment  is  subject  to  an  upward 
force  or  clockwise  couple  respectively,  and  vice  versa. 

Under  normal  stress  a  body  simply  tends  to  lengthen  or 
shorten  (according  as  the  stress  is  compression  or  tension) 
along  a  line  at  right  angles  to  the  section.  Compression  is 
the  stress  typical  of  columns,  posts,  struts,  and  pedestals. 
Tension  is  typical  of  ropes  and  chains  in  service,, or  tie-rods. 
Anything  through  which  a  push  is  transmitted  is  in  compres- 
sion, and  anything  through  which  a  pull  is  transmitted  is  in 
tension. 

In  Shear  one  segment  of  the  body  tends  to  slide  by  the 
other  with  no  tendency  to  rotate  about  an  axis  normal  to  the 
section.  Shear  is  the  stress  typical  of  rivets;  in  fact,  their 
main  purpose  is  usually  to  resist  this  kind  of  stress.  The 
shearing  resistance  of  the  rivet  in  Fig.  20  prevents  it  from 


FIG.  20. 

being  separated  into  three  parts  by  shearing  on  the  planes  of 
contact  of  the  plates  which  it  connects.  Another  illustration 
of  shear  is  to  be  found  (Fig.  21)  in  the  plane  indicated  by 
the  dotted  line  in  the  timber  receiving  thrust  from  the  rafter. 

Flexure  is  simply  a  tendency  to  bend.  Familiar  examples 
are  a  stick  bent  over  the  knee,  or  a  loaded  floor  joist. 

Torsion  is  a  twisting  tendency  due  to  the  two  segments  of 


82  STATICS. 

the  body  tending  to  rotate  in  opposite  directions  about  an  axis 
normal  to  the  section.  Torsion  is  the  stress  typical  of  shaft- 
ing for  the  transmission  of  power.  It  is  rarely  permitted  else- 
where if  it  can  be  avoided. 

When  it  is  said  that  Flexure  or  Torsion  exists  at  a  given 
section  it  means  that  the  segment  of  the  body  on  one  side  of 


FIG.  21. 

the  section  is  likely  to  bend  or  twist  respectively  with  respect 
to  the  other  segment.  These  two  stresses  may  be  looked  upon 
as  complex  cases  of  normal  stress  and  shear  respectively,  i.e. 
cases  in  which  the  tendency  to  failure  under  the  normal  stress 
or  shear  varies  in  different  parts  of  the  section.  It  is  familiar 
fact  that  the  upper  and  lower  surfaces  of  a  bent  beam  and  the 
surface  portions  of  a  shaft  are  subject  to  severer  stress  than  the 
interior  portions. 

Familiar  examples  of  combined  stresses  occur  in  beams 
subject  to  transverse  loads  in  which  shear  and  flexure  appear 
together;  in  shafting  in  which  shear,  flexure,  and  torsion  ap- 
pear together,  to  which  may  be  added  compression  if  the  shaft 
is  vertical. 

A  stress  is  fully  described  as  soon  as  its  magnitude  and 
nature  are  stated,  nature  being  here  understood  to  include 
algebraic  sign  as  well  as  kind  (§  59). 


CHAPTER   X. 
STRUCTURES. 

62.  Structures.  Definitions.— Structures  are  simply  artifi- 
cial contrivances  for  supporting  loads  or  resisting-  the  active 
forces  of  nature.  They  must  resist  the  stresses  due  to  the 
action  of  these  loads  or  forces  on  the  one  hand  and  the  reaction 
of  the  earth's  surface  on  the  other.  They  may  be  very  simple 
(as  mere  posts,  pedestals,  tension  rods),  acting  under  com- 
pression alone  or  under  tension  alone,  and  such  may  be  desig- 
nated by  the  term  elemental  structures.  In  general,  however, 
the  term  structure  will  be  understood  to  mean  only  the 
complex  kind,  such  as  are  subject  as  a  whole  to  bending  and 
shear. 

As  regards  their  composition,  structures  may  be  divided 
into  two  broad  types,  (a)  framed,  and  (b)  non-framed. 

A  framed  structure,  or  frame,  is  one  composed  of  a  series 
of  straight  bars  fastened  together  by  their  ends  only,  so  as  a 
whole  to  make  substantially  one  rigid  body. 

Ideally  (that  is,  if  the  joints  could  be  made  frictionless 
hinges,  if  the  weights  of  the  bars  could  be  made  to  act  only 
at  their  ends,  and  if  all  other  loads  are  applied  only  at  the 
joints)  the  stress  in  each  bar  of  a  frame  would  be  purely  axial 
(i.e.,  pure  tension  or  pure  compression),  making  each  member 
an  elemental  structure. 

Such  ideal  frames  are  also  called  true  frames,  and  are  what 
is  meant  when  frames  are  referred  to  without  further  description. 

The  nearest  actual  approximation  to  a  true  frame  is  a  pin- 

83 


84  STATICS. 

connected  truss.  Indeed  pin-connections  have  been  used  very 
widely  in  truss  construction  largely  because  the  assumed  condi- 
tions of  computation  can  thus  be  most  nearly  realized  in  prac- 
tice (cf.  §  63). 

The  resultant  of  the  external  forces  at  any  section  of  a  true 
frame  is  resisted  by  elemental  members  into  which  the  section 
can  easily  be  analyzed  (for  the  center  lines  of  the  bars  must 
be  the  lines  of  action  of  the  forces  internal  to  the  structure)  and 
each  pin  or  joint  is  in  equilibrium  under  the  action  of  a  set  of 
concurrent  forces. 

A  non-framed  structure  is  one  consisting  of  one  continu- 
ous member  or  of  a  number  of  members  so  fastened  together 
throughout  their  lengths  as  to  make  one  solid  piece. 

A  non-framed  structure  cannot  be  analyzed  into  separate 
elemental  members.  The  flexure  and  shear  at  any  section  is 
resisted  by  the  whole  section  under  stress  varying  in  intensity 
from  point  to  point  of  the  section  according  to  more  or  less 
complex  laws. 

The  relations  between  the  external  forces  are  entirely  in- 
dependent of  whether  the  structure  is  framed  or  non-framed. 
The  difference  in  the  types  is  wholly  one  of  internal  make-up, 
and  effects  the  method  of  dealing  with  internal  forces  only. 

The  periphery  of  a  frame  constitutes  what  are  called  the 
two  chords  of  the  frame,  the  portion  of  the  periphery  bounded 
by  the  end  joints  on  the  upper  side  of  the  frame  being  further 
designated  as  the  top,  or  upper,  chord  and  the  one  on  the 
lower  side  as  the  bottom,  or  lower,  chord. 

63.  Extent  of  Approximation  to  True  Frames  in  Practice. 
— Outside  of  the  fact  that  joints  cannot  possibly  be  made  fric- 
tionless,  and  that  the  weights  of  the  bars  cannot  possibly  be 
made  to  act  only  at  their  ends,  true  frames  are  very  uncommon. 

Compression  chords,  that  is,  the  large  part  of  the  periphery 
of  every  frame  which  is  subject  to  compression,  are  usually 


STRUCTURES.  85 

0 

made  stiff  at  the  joints.  Other  members  may  be  hinged  to 
them,  but  the  chords  themselves  are  not  broken  at  the  joints, 
or  are  so  rigidly  spliced  as  to  be  practically  continuous  and 
hence  stiff.  Moreover,  trusses  are  frequently  made  with  no 
hinge  joints  whatever,  but  with  tightly  riveted  connections. 

The  conception  of  the  true  or  ideal  frame  as  above  defined 
is,  however,  fundamental  to  the  design  of  all  these  structures, 
and  is  the  hypothesis  upon  which  the  normal  stresses  in  the 
members  are  determined.  The  difference  between  this  ideal 
frame  and  actual  ones  can,  by  proper  design  and  construction, 
be  rendered  negligible.  When,  however,  the  difference  is 
permitted  to  be  considerable,  means  are  available  for  estimat- 
ing and  providing  for  the  so-called  secondary  stresses  thereby 
produced. 

If  the  axes  of  the  various  bars  intersect  at  common  points, 
i.e.,  if  the  forces  at  a  joint  are  really  concurrent,  secondary 
stresses  are  considered  negligible  in  spite  of  the  joint  being  far 
from  frictionless.  This  is  due  to  the  materials  being  so  nearly 
rigid  that  only  very  slight  changes  of  shape  in  the  structure 
occur  and  hence  there  occurs  only  a  very  slight  tendency  for 
the  bars  to  turn  about  the  joints.  The  forces  themselves  by 
meeting  at  the  joint  produce  no  tendency  of  this  kind. 

64.  Loads  Applied  Elsewhere  than  at  Joints. — It  occa- 
sionally happens  that  loads  must  be  brought  to  bear  on  frames 
at  points  where  there  is  no  joint,  and  where  it  is  not  practicable 
to  make  one  by  the  addition  of  more  bars  to  the  frame.  The 
result  is  that  the  bar  to  which  the  force  is  applied  has  to  do 
double  duty — that  of  any  frame  member  subject  to  tension  or 
compression  and  also  that  of  a  beam  or  girder  subject  to  shear 
and  flexure. 

In  its  capacity  as  a  beam  the  bar  transmits  its  transverse 
loads  to  the  frame  in  the  form  of  reactions  upon  the  joints  by 
which  the  bar  is  incorporated  into  the  frame.  These  reactions 


86 


STATICS. 


are  determined  as  in  any  case  of  a  beam  with  given  loads  and 
supports.  Reversing  their  senses,  they  are  the  forces  acting 
on  the  joints  of  the  frame.  Making  these  determinations  and 
substitutions  the  frame  can  be  analyzed  as  usual.  The  shear 
and  flexure  for  which  (in  addition  to  their  tension  or  compres- 
sion as  part  of  the  frame)  the  bars  subject  to  transverse  loads 
must  be  designed  are  determined  as  for  any  beam. 

This  form  ot  construction  is  uneconomical  of  material,  and 
is  usually  to  be  avoided. 

Strictly  speaking,  all  bars  of  a  material  frame  are  in  the 
condition  just  described  owing  to  their  own  weight,  but  in 
frames  of  moderate  size  the  resulting  shear  and  flexure  in  indi- 
vidual bars  is  neglected  even  though  the  frame  itself  is  analyzed 
for  tensions  and  compressions  due  to  its  own  weight  considered 
concentrated  at  the  joints. 

For  example  of  a  frame  illustrating  this  section  see  §  84 
and  Exercise  27. 

65.  Frames  in  General. — Frames  may  be  (a)  complete, 
(£)  incomplete,  or  (c)  redundant.  A  complete  frame  is  one 
composed  of  just  enough  bars  to  insure  its  keeping  its  shape 
under  all  conditions  of  loading.  If  it  has  fewer  bars  than  this 


FIG.  24. 

requires,  but  nevertheless  is  able  to  carry  a  load  if  properly 
distributed,  it  is  incomplete,  and  if  more,  it  is  redundant. 
Examples  of  each  are  shown  in  Figs.  22,  23,  and  24  re- 
spectively. 


STRUCTURES.  87 

Since  the  triangle  is  the  only  geometrical  figure  in  which 
a  change  of  shape  is  precluded  unless  the  lengths  of  its  sides 
are  changed,  the  triangle  is  necessarily  the  basis  of  arrange- 
ment of  the  bars  in  a  frame, 

A  complete  frame  must  accordingly  be  made  up  of  the 
minimum  number  of  bars  consistent  with  its  being  composed 
wholly  of  triangles.  Removing  a  bar  common  to  two  triangles 
would  render  it  incomplete.  Adding  bars  to  a  complete  frame, 
as  by  adding  the  second  diagonal  in  one  or  more  quadrilaterals, 
renders  it  redundant,  but  if  such  bars  are  capable  of  resisting 
one  kind  of  stress  only,  as  in  the  case  of  counters,  the  redun- 
dancy may  be  only  apparent.  See  §  83  and  Exercise  26. 

Complete  frames  are  the  type  which  is  usually  closely  ap- 
proximated, and  which  accordingly  receives  most  attention. 

Their  analysis  involves  only  a  straightforward  application 
of  the  familiar  principles  of  statics. 

Incomplete  frames  are  stable  only  under  symmetrical  or 
other  specially  arranged  loads.  Under  such  loads  they  are 
analyzed  with  as  much  ease  and  certainty  as  Complete 
Frames,  and  require  no  further  explanation. 

Structures  having  outlines  of  incomplete  frames  may  resist 
loads  of  any  distribution  by  virtue  of  the  flexural  strength  of 
members  continuous  through  several  joints,  but  such  structures 
are  not  true  frames  and  require  special  treatment,  which  takes 
note  of  the  ability  of  some  of  their  bars  to  resist  bending. 

Redundant  frames  *  will  resist  loads  of  any  distribution,  and 
some  forms  are  not  uncommonly  found  advantageous  in  use. 
Their  analysis  involves  statically  indeterminate  problems  and 

*  A  test  for  redundancy  can  be  worked  out  as  follows.  Beginning  with  a  tri- 
angle, each  two  bars  added  establishes  a  new  joint.  Then  if  3  -{-  x  equals  the 
number  of  joints,  the  number  of  bars  for  the  complete  structure  will  be  3  -j-  2x,  i.e., 
twice  the  number  of  joints  minus  3.  Therefore,  for  complete  frames,  if  m  equals 
the  number  of  bars  and  n equals  the  number  of  joints,  m  equals  2«— 3.  Km  is 
less  than  (2«— 3),  the  frame  is  incomplete;  if  m  is  more  than  (2^—3),  the  frame 
is  redundant. 


88  STATICS. 

their  stresses  therefore  cannot  be  worked  out  by  purely  statical 
methods,  but  with  special  data  as  to  the  material,  or  as  to  the 
perfectness  of  workmanship,  or  by  the  aid  of  some  outright 
assumption  as  to  probable  action  of  the  bars,  a  more  or  less 
satisfactory  estimate  of  stresses  can  be  made. 

66.  Loads. — Loads  are  (a)  permanent  or  dead,  those  which 
are  always  present;  or  (b)   moving  or  live,  those  which  are 
only  occasionally  present. 

The  permanent  load  for  roofs  is  the  weight  of  the  roof  cov- 
ering, purlins,  trusses,  etc. 

The  live  load  for  roofs  is  wind  pressure,  snow,  etc. 

The  permanent  load  on  bridges  is  the  weight  of  the  floor, 
trusses,  etc. 

The  live  load  on  bridges  is  the  weight  of  trains,  carts, 
crowds  of  people,  wind  pressure,  etc. 

Loads  which  act  simultaneously  on  a  structure  may  or  may 
not  be  considered  all  at  once  in  the  determination  of  stresses. 
If  they  are  not  considered  all  at  once  the  total  effect  is  ob- 
tained by  simply  taking  the  algebraic  sum  of  all  the  stresses 
caused  by  the  partial  loads.  It  is  usually  desirable  in  practice 
to  follow  this  course  for  the  sake  of  avoiding  very  serious  com- 
plications in  the  work. 

The  way  in  which  stresses  are  provided  for  falls  within  the 
domain  of  Resistance  of  Materials. 

67.  Stresses  in  Structures. — As  has  been  stated,  stresses 
result  directly  from  external  forces.      External  forces  are  either 
loads  or  reactions.      Loads  are  always  known  or  assumed  and 
the  reactions  determined  accordingly  by  the  methods  already 
developed,  commonly  Case  2b  or  Case  3.* 

*  It  may  be  noted  that  certain  cases  of  statically  indeterminate  problems  occa- 
sionally arise  in  connection  with  reactions.  An  example  would  be  found  in  a 
beam  resting  on  three  or  more  supports,  giving  rise  to  three  or  more  parallel 
reactions.  The  reactions  can  be  worked  out  in  such  cases,  by  the  aid  of  the  laws 
of  elasticity,  involving  methods  outside  the  scope  of  this  book.  The  need  of  re- 
sorting to  such  methods  is  usually  avoided  in  the  design  of  the  structure. 


STRUCTURES.  89 

As  soon  as  the  reactions  are  known,  the  external  forces  are 
all  known  and  the  stresses  can  be  determined  as  a  purely  statical 
matter  by  methods  to  be  inferred  from  the  discussion  of  stresses 
in  §§  58-60.  As  there  stated  the  stress  at  any  section  is  really 
a  question  of  the  resultant  of  all  the  forces  external  to  the 
body  on  one  of  the  segments  into  which  the  section  divides  the 
body.  This  resultant  is  a  measure  of  what  may  be  looked 
upon  as  normal  stress,  shear,  flexure,  or  torsion  whether 
the  body  is  framed  or  non-framed.  Before  either  kind  o 
structure  can  be  designed,  however,  the  statical  analysis  must 
go  a  step  further  still.  In  the  case  of  non -framed  structures 
this  further  analysis  is  statically  indeterminate  and  requires 
the  aid  of  the  elementary  principles  of  Resistance  of  Materials 
for  its  accomplishment.  Stresses  in  a  non-framed  structure  are 
accordingly  considered  determined  as  soon  as  each  of  the  four 
kinds  of  stress  is  fully  known  at  a  requisite  number  of  sections. 

In  framed  structures  this  further  analysis  may  be  statically 
determinate.  It  consists  in  determining  just  how  much  tension 
or  compression  exists  in  each  bar  as  a  consequence  of  a  given  set 
of  loads.  The  determination  of  stresses  in  a  framed  structure 
does  not  stop  with  the  determination  of  normal  stress,  shears, 
flexures,  and  torsions  in  the  body  as  a  whole.  In  fact,  in 
many  cases  it  proceeds  directly  to  the  determinations  of  the 
tensions  and  compressions  in  the  bars  consequent  upon  the  four 
fundamental  stresses  without  stopping  to  find  tnose  stresses  in 
their  unanalyzed  state  at  all. 

The  next  two  chapters  will  state  the  statical  processes  ap- 
plied to  each  of  the  two  kinds  of  structures  after  the  external 
forces  are  all  known. 


CHAPTER   XL 
STRESSES   IN  NON-FRAMED  STRUCTURES. 

68.  Stresses  in  Non-framed  Structures. — In  a  non-framed 
structure,  stresses  are  found  at  any  section  by  finding  the  re- 
sultant of  all  the  forces  on  one  side  of  the  section.  Its  com- 
ponents normal  and  parallel  to  the  section  will  measure  the 
normal  stress  and  shear  respectively,  and  the  senses  of  the 
components  will  decide  whether  the  former  is  compression  or 
tension  and  the  latter  positive  or  negative.  Taking  the  inter- 
section of  the  force  with  the  section  plane  as  its  point  of  ap- 
plication, the  moment  of  the  normal  component  and  of  the 
parallel  component  about  the  center  of  gravity  will  measure 
the  flexure  and  torsion  respectively,  and  the  signs  of  the  mo- 
ments will  distinguish  between  positive  and  negative  values  of 
the  stresses.  This  process  is  simply  Cases  I  and  2.a  combined. 

A  process  statically  identical  with  the  preceding  is  fre- 
quently more  convenient  in  practice.  In  this  process  the  com- 
ponents of  the  individual  forces  normal  and  parallel  to  the 
section  are  summed  to  get  the  normal  stress  and  shear  respec- 
tively, and  the  moments  of  the  individual  forces  about  the 
center  of  gravity  of  the  section  and  about  an  axis  through  the 
center  of  gravity  normal  to  the  section  are  summed  to  get  the 
flexure  and  torsion  respectively. 

In  non-framed  structures,  stresses  have  to  be  worked  out 
at  only  a  relatively  small  number  of  critical  sections.  It  is 

sufficient  at  other  sections  to  be  sure  that  the  stresses  do  not 

90 


STRESSES  IN  NON-FRAMED  STRUCTURES.  91 

exceed  certain  amounts.  Stresses  existing  at  all  sections  can 
easily  be  shown,  if  required,  as  ordinates  of  properly  con- 
structed curves.  (Cf.  Exercise  18.) 

Exercise  18.  A  horizontal  bar  is  acted  on  by  (20  240°  5,  o),  (30  300° 
8,  o),  (40  90°  20,0),  and  by  a  force  at  (o,  o)  whose  line  of  action  is  inclined 
30°  to  the  horizon,  and  one  at  (12,  o)  unknown  both  in  magnitude  and 
direction.  Establish  equilibrium  and  determine  all  the  stresses  at  a 
section  through  (10,  o).  Solve  graphically  only.  Show  clearly  on  the 
drawing  how  results  were  found  and  where  they  were  scaled. 

69.  Shear  Diagrams. — A  diagram  showing  the  value  of 
the  shear  at  all  points  of  a  vertically  loaded  beam  can  easily 
be  constructed  by  selecting  a  horizontal  base  line  and  drawing 
parallels  across  the  intervals  between  the  adjacent  forces  at  dis- 
tances above  or  below  the  base  line  proportional  to  the  magni- 
tude of  the  sum  of  the  forces  on  either  side  of  the  interval. 
Resultants  of  forces  at  the  left  of  any  segment  indicate  positive 
or  negative  shears  and  are  set  off  upward  or  downward  from 
the  base  line  according  as  they  are  upward  or  downward. 

Thus,  in  Fig.  25,  the  values  of  the  shears  in  proceeding 
from  left  to  right  are  in  the  four  intervals  respectively,  +  5 l  -77> 
—48.23,  —18.23,  and  +31-77-  The  shear  diagram  is  con- 
sequently as  shown  in  the  lower  shaded  diagram. 

The  four  numerical  values  just  given  are  evidently  the  values 
of  the  resultants  AB,  AC,  AD,  and  AE,  and  could  have  been 
projected  across  into  their  respective  intervals  from  the  magni- 
tude diagram. 

70.  Flexure  Diagrams. — Since  the  measure  of  flexure  at 
any  section  is  the  sum  of  the  moments  about  the  section  of  all 
the  forces  on  one  side  of  the  section,  all  that  is  necessary  for 
obtaining  a  diagram  showing  the  flexure  at  all  sections  of  a  beam 
under  a  set  of  parallel  forces  is  to  letter  the  forces  in  the  order 
of  their  occurrence  along  the  beam  and  draw  their  closed  string 
polygon  (§  52).      The  intercept  by  this  polygon  from  a  line 
parallel  to  the  forces  is  proportional  to  the  flexure  in  the  sec- 


92  STATICS. 

tion  of  the  beam  traversed  by  that  line.  The  intercept  needs 
only  to  be  multiplied  by  the  common  H  (§  5  i)of  the  forces  to 
give  the  numerical  value  of  that  flexure. 

An  example  of  a  flexure  diagram  for  a  beam  subject  to  a 
series  of  vertical  forces  is  shown  in  the  upper  of  the  two  shaded 
diagrams  in  Fig.  25.  To  find  the  flexure  at  any  section  all 


31.77 


FIG.  25. 

that  is  needed  is  to  drop  a  vertical  line  from  that  section,  meas- 
ure the  portion  of  this  vertical  within  the  diagram  and  multiply 
the  resulting  length  by  30  Ibs.,  which  was  selected  as  a  con- 
venient value  for  H  in  constructing  the  diagram. 


STRESSES  IN  NON-FRAMED  STRUCTURES.  93 

p! 

Thus  the  intercept  directly  below  the  load  be  scales  5.2 
ft.  The  flexure  in  the  section  directly  under  be  is  therefore 
5.2  X  30  =  I56ft.-lbs.  Algebraically  the  value  is  found  to 
be  more  precisely  51.77X3  —  15  5-  31  ft.  Ibs.  Values  can  be 
obtained  in  like  manner  for  any  other  section.  It  should  be 
observed  that  the  percentage  accuracy  of  the  graphical  result 
is  no  greater  than  the  accuracy  of  the  scaling  of  the  intercept. 
With  small  intercepts  therefore  the  percentage  of  inaccuracy  in 
the  graphical  result  may  be  considerable. 

Intercepts  above  ao  indicate  positive  flexure  and  those 
below  ao  negative  flexure. 

Observe  that  the  diagram  shows  at  a  glance  that  the  flex- 
ure in  the  beam  increases  steadily  to  a  positive  maximum  under 
be,  then  decreases  to  zero  at  a  point  a  little  nearer  to  cd  than 
to  be  and  continues  to  increase  negatively,  though  more  slowly, 
after  passing  cd  to  a  negative  maximum  under  de,  when  it  neg- 
atively decreases  rapidly  again  to  zero  under  ea.  The  nega- 
tive maximum  is  found  to  be  numerically  slightly  larger  than 
the  positive  maximum,  and  the  dangerous  section  of  the  beam, 
so  far  as  flexure  is  concerned,  is  under  de. 

Observe  that  the  sections  where  the  flexure  reaches  its 
greatest  values,  whether  positive  or  negative,  are  those  in 
which  the  shear  passes  through  zero — a  phenomenon  of  in- 
evitable occurrence,  as  will  be  shown  in  the  next  section. 

Exercise  19.  A  horizontal  beam  of  30  ft.  span,  supported  at  each  end, 
carries  loads  of  300,  600,  1800,  and  1200  Ibs.  at  points  6,  10,  18,  and  25  ft. 
respectively  from  the  left  end.  Neglecting  the  weight  of  the  beam  itself, 
determine  by  both  methods  the  numerical  value  of  the  flexure  at  sec- 
tions 8,  12,  1 8,  and  30  ft.  from  the  left  end.  Record  results  side  by  side 
for  comparison. 

Suggestions.  Take  the  scale  of  lengths  as  great  as  i  in.  =  4  ft.  Take 
some  convenient  round  number  for  the  magnitude  of  H. 

71.  Connection  Between  Shear  and  Change  in  Flexure. — 

It  will  be  useful  to  see  if  there  is  a  simple  relation  between  the 


94  STATICS. 

flexure  at  the  end  of  an  interval  and  the  shear  and  flexure  at 
the  beginning  of  the  interval. 

Accordingly,  let  AC,  Fig.  26, 

be  any  segment  of  a  beam,  Cl  and 
c2c    £^  and  j^  and  j^  the  ends  of  the 

_C!          interval  and  values  of  the  flexures 
FIG.  26.  existing  there  respectively,  PL  the 

resultant  of  all  the  forces  at  the  left 

of  Cv  P2  of  those  in  the  interval  C^CZ.     Dimensions  as  shown. 
Then 


and  Ml  =  P^ar 

Subtracting  the  latter  from  the  former, 
Mt-Jf1  =  Pl  (a,  +  as) 
But  Pl  is  the  measure  of  the  shear  at  the  beginning  of  the 
interval  and  (a2  -\-  az)  is  the  length  of  the  interval.      Calling 
the  former  V,  the  latter  a,  giving  P2a3  the  more  convenient 
characterization  -f-  My  the  last  equation  takes  the  form 

Mt  =  M^  +  Va  +  Mr 

That  is,  the  flexure  at  the  end  of  an  interval  is  measured 
by  the  algebraic  sum  of  the  flexure  at  the  beginning  of  the 
interval,  the  product  of  the  shear  at  that  section  by  the 
length  of  the  interval,  and  the  sum  of  the  moments  of  the 
forces  in  the  interval  about  the  center  of  gravity  of  the  end 
of  the  interval.* 

Corollary.  If  a  be  of  infinitesimal  length,  dx,  M3  will  van- 
ish, M2  —  Ml  may  be  written  dM,  and  the  equation  takes  the 
form 

dM 
V  ~  dx  ' 

*  The  data  of  Fig.  26  are  of  the  simple  sort  usual  in  the  practical  examples  of 
this  problem.  The  reader  should  satisfy  himself  that  the  same  conclusion  would 
have  been  reached  whatever  the  directions  of  PI  and  /*,,  or  if  one  or  both  of  them 
had  been  couples. 


STRESSES  IN  NON-FRAMED  STRUCTURES.  95 

0 

wnence  it  appears  that  the  shear  at  any  section  is  the  ^-deriva- 
tive ot  the  flexure  at  that  section,  and  it  follows  that  where 
the  shear  passes  through  zero  the  flexure  is  at  a  maximum 
or  minimum. 

Exercise  20.  Shears  and  Flexures  from  fixed  loads.  Draw  diagrams 
of  shears  and  flexures  for  each  of  the  seven  following  cases.* 

(a)  Cantilever  with  W  concentrated  at  the  outer  end ;  (b)  cantilever 
with  W  uniformly  distributed;  (c)  simple  beam  with  ^concentrated  at 
the  middle ;  (//)  simple  beam  with  W  uniformly  distributed  ;  (e)  simple 
beam  under  several  concentrated  loads ;  (_/")  beam  overhanging  one 
support,  concentrated  load  at  the  end  of  the  overhang,  two  others  be- 
tween the  supports ;  (g]  beam  with  equal  loads  at  each  end,  supports  at 
equal  distances  from  the  ends. 

Compute  algebraically  and  compare  the  maximum  shears  and  flex- 
ures in  cases  (a),  (b),  (<:),  (c/). 

Suggestions.  Note  that  in  (a),  (c\  and  (e)  the  diagram  of  shear  can  be 
transferred  directly  from  the  magnitude  diagram, — strictly  also  in  (b) 
and  (d). 

In  dealing  with  (b}  and  (d]  construct  approximate  diagrams  by  divid- 
ing the  load  into  short  portions,  and  treat  as  a  set  of  equivalent  loads 
concentrated  at  the  centers  of  gravity  of  these  portions..  Points  in  the 
true  diagrams  will  lie  at  the  intersections  of  the  extended  vertical 
boundaries  of  the  portions  with  the  approximate  diagrams  thus  found 
and  in  the  case  of  shear  will  be  the  straight  line  through  these  points; 
in  the  case  of  flexure  will  be  the  parabola  inscribed  in  the  approximate 
string  polygon  and  tangent  to  it  at  these  points. 

*  Here  the  term  cantilever  is  used  to  designate  a  beam  supported  at  one  end 
only  (by  being  built  rigidly  into  a  wall,  for  example)  or  in  any  way  overhanging  a 
support ;  a  simple  beam  is  understood  to  mean  one  resting  upon  supports  at  each 
end  without  constraint. 


CHAPTER   XII. 

STRESSES   IN   FRAMED    STRUCTURES. 

72.  Stresses  in  Framed  Structures. — In  a  framed  structure 
the  resultant  of  all  the  loads  on  one  side  of  a  section  can  pass 
the  section,  so  as  to  hold  the  other  segment  in  equilibrium, 
only  in  the  form  of  simple  pushes  and  pulls  which  must  act 
along  the  center  lines  of  the  bars  cut.  These  bars  should  be 
imagined  to  be  replaced  by  forces  acting  along  their  center 
lines,  which  are  a  set  of  forces  into  which  the  resultant  can  be 
resolved — a  problem  which  is  determinate  if  the  number  of 
forces  does  not  exceed  two  or  three.  This  resolution  accom- 
plished (Case  2  or  Case  4),  the  compression  or  tension  in 
each  bar  is  known  and  the  requisite  stresses  determined. 

Here  as  with  non-framed  structures  it  is  usually  unneces- 
sary actually  to  evaluate  the  resultant  of  the  external  forces 
on  one  side  of  the  section.  The  individual  external  forces  are 
treated  as  the  given  forces  in  Case  2  or  Case  4,  and  the  un- 
knowns are  the  two  or  three  components  of  the  resultant  which 
measure  the  required  stresses. 

This  method  could  be  applied  repeatedly  until  every  bar 
in  the  structure  had  been  cut  and  the  stress  in  it  determined, 
but  when  the  stresses  in  a  large  number  or  in  all  of  the  bars  in 
the  frame  are  to  be  found,  a  less  laborious  method  suggests 
itself,  when  it  is  observed  that  each  and  every  joint  is  in  equi- 
librium under  a  set  of  concurrent  forces.  Selecting  a  joint  if 
possible  where  only  two  bars  concur  with  one  or  more  external 
forces,  the  stress  in  these  two  bars  can  be  found  by  Case  2#. 

96 


STRESSES  IN  FRAMED  STRUCTURES.  97 

0 

The  joint  at  the  other  end  of  one  of  these  bars  will  then  com- 
monly be  found  to  be  subject  to  forces,  only  two  of  which  are 
now  unknown — one  of  those  previously  unknown  being  the 
one  just  established,  for  the  force  which  a  bar  exerts  upon  a 
joint  at  one  of  its  ends  must  be  the  equal  and  opposite  of  that 
exerted  upon  the  joint  at  the  other  end.  Proceeding  thus, 
applying  Case  2a  to  joint  after  joint  till  all  have  been  treated, 
the  stresses  in  all  the  bars  are  known. 

If  there  are  no  joints  where  so  few  as  two  bars  concur,  the 
problem  is  of  course  statically  indeterminate  unless  the  frame 
is  so  composed  as  to  permit  the  determination  of  one  of  three 
concurring  bars  by  using  method  of  sections,  §  73.  (Cf. 
Exercise  25.) 

Sometimes  a  similar  step  has  to  be  taken  to  deal  with  joints 
in  the  interior  frames  of  where  three  forces  concur,  all  incapable 
of  determination  by  the  ordinary  method  of  Case  2a.  (Cf. 
Exercise  24.) 

73.  Method  of  Sections. — Method  of  Sections  *  is  the  name 
applied  to  the  method  of  determining  stresses  in  a  framework 
by  dividing  the  structure  into  two  segments  by  means  of  a  sec- 
tion plane,  treating  the  bars  cut  as  mere  lines  of  action  of 
forces  external  to  the  segment,  and  rinding  magnitudes  accord- 
ingly. The  problem  presented  is  Case  4,  when  three  bars  are 
cut  by  the  section,  or  Case  2,  in  the  rarer  case,  when  only  two 
are  cut.  The  calculations  can  be  made  for  whichever  segment 
the  work  will  be  easier,  or,  if  a  check  be  important,  for  both 
segments. 

*  This  method  solved  algebraically  is  sometimes  more  explicitly  called  Ritter's 
Method  of  Sections,  after  Professor  August  Ritter,  of  Aix-la-Chapelle,  who  used  it 
freely  in  his  Dach-  u,  Briicken-Constructionen.  Solved  graphically  the  process 
is  also  called  Culmann's  Method  of  Sections,  after  Professor  Culmann,  of  Zurich. 

It  must  be  observed  that  the  methods  used  in  this  book  for  finding  stresses  are 
really  all  methods  of  sections,  and  the  limitations  imposed  in  this  section  must  be 
seen  to  be  arbitrary  or  conventional,  with  a  view  to  the  establishment  of  a  con- 
venient  technical  term. 


98 


STATICS. 


Wherever  we  can  divide  a  structure  without  cutting  more 
than  three  bars,  if  these  three  bars  do  not  meet  in  a  point,  the 
stresses  in  the  bars  can  be  determined. 

If  stress  in  a  single  bar  only  is  desired,  one  simply  selects 
a  convenient  section  plane  which  will  cut  this  bar  and  only  two 
other  bars. 

The  algebraic  method  of  solving  Case  4  will  usually  be 
found  more  convenient  than  the  graphic  for  these  problems. 

Lever-arms,  however,  are  sometimes  so  troublesgme  to 
calculate  that  scaling  them  from  a  carefully  made  large  scale 
drawing  is  the  best  way  to  get  them.  Writing  moments  in 
form  P  ( y  cos  of  —  _rsin  a)  is  always  an  available  resource. 

The  reasoning  underlying  this  method  is  still  further  ex- 
plained by  Fig.  27.  Fig.  2/#  represents  any  frame  in  equilib- 


(a) 


FIG.  27. 

rium  under  the  external  forces  P,  Q,  R,  5,  and  T.  It  is  re- 
quired to  find  the  stress  in  V.  Intersecting  the  frame  by  a 
section  cutting  V  and  two  other  bars,  U  and  Wy  we  have  as  a 


STRESSES  IN  FRAMED  STRUCTURES.  99 

result  what  can  be  regarded  as  two  rigid  bodies  shown  shaded 
in  Figs.  27 b  and  27 'c  in  equilibrium  under  the  forces  (external 
to  them)  T,P,Q,  U,  V,WmA  R,S,C7,  V,  W,  in  either  of  which 
V  can  be  established  by  an  equation  of  moments  with  Mv, 
the  intersection  of  U  and  W,  as  a  center.  If  stresses  in  U 
and  J^be  also  required,  they  can  be  found  by  taking  centers  at 
M u  and  M  m  respectively. 

Observe  that  this  solution  is  entirely  independent  of  the 
number  or  inclination  of  bars  in  the  frame  other  than  those  cut, 
provided  P,  Q,  R,  5,  and  T  remain  unchanged  in  all  their 
elements. 

Plate  V  may  be  looked  upon  as  giving  a  complete  graphic 
as  well  as  algebraic  solution  by  this  method,  if  the  shaded 
body  be  considered  the  segment  of  a  frame,  and  the  three 
forces,  P,  Q,  and  R,  the  forces  acting  through  the  cut  bars. 

74.  Method  for  Determining  All  the  Stresses  in  a  Frame 
Under  a  Given  Load. — To  determine  all  the  stresses  in  a  non- 
redundant  frame,  since  it  is  made  up  of  a  series  of  sets  of  con- 
current forces,  we  need  only  to 

(1)  See  that  the  external  forces  are  in  equilibrium,  and 

(2)  Work  out  case  2a  for  each  joint. 

In  the  former  of  these  two  processes  the  algebraic  method 
is  generally  to  be  preferred.  It  usually  is  less  laborious,  and 
the  superior  precision  of  its  results  is  welcome,  as  the  accuracy 
of  all  the  succeeding  work  depends  upon  them. 

In  the  latter  process,  however,  the  graphic  method  is 
greatly  to  be  preferred,  as  saving  much  troublesome  labor  with 
a  minimum  risk  of  serious  error  with  a  degree  of  accuracy 
amply  sufficient  for  the  needs  of  engineering  design.  This 
work  consists  simply  of  beginning  with  a  joint  where  two  bars 
concur  with  one  or  more  external  forces,  and  working  out  a 
closed  magnitude  polygon  for  the  set  of  concurrent  forces  so 
composed.  Two  of  the  sides  of  this  polygon  will  determine 


ioo  STATICS. 

the  stresses  m  the  two  bars — not  only  in  magnitude,  but  also 
in  nature,  the  former  being  shown  by  the  lengths  of  the  sides, 
the  latter  by  the  arrows  upon  them,  for  these  arrows  must 
follow  one  another  around  the  polygon,  and  when  transferred 
to  the  bars  to  which  they  belong  they  indicate  compression  or 
tension  according  as  they  point  toward  or  away  from  the  joint. 
Proceeding  to  the  next  joint,  which,  with  the  force  acting 
through  the  first  two  bars  known,  will  in  turn  have  only  two 
unknowns  concurring  upon  it,  a  second  polygon  is  constructed 
and  so  on  through  the  whole  series  of  joints.  Thus  would 
result  a  series  of  magnitude  polygons,  one  for  each  joint. 

75.  Example. — An  example  is  worked  out  in  Fig.  28. 
The  frame  there  shown,  Fig.  2$>a,  is  known  to  be  in  equilib- 
rium under  the  five  external  forces,  20,  200,  30,  140,  and 
no  Ibs.',  through  the  first  three  having  been  given  at  the 
outset  and  the  last  two  having  been  determined  (Case  2,b) 
from  the  dimensions  of  the  frame  and  the  positions  of  the  sup- 
ports. 

It  is  most  convenient  to  letter  the  external  forces  in  the 
order  of  their  occurrence  around  the  outside  of  the  figure  and 
have  the  letter  common  to  two  adjacent  forces  apply  also  to 
the  bar  or  bars  in  the  periphery  of  the  frame  between  their 
points  of  application.  Letters  added  inside  each  triangle  of 
the  frame  complete  the  lettering  of  the  forces. 

The  work  can  begin  at  either  of  the  two  end  joints.  Tak- 
ing the  left  end  joint,  we  lay  off  EAB  as  the  beginning  of  its 
magnitude  polygon;  then  a  line  from  B  parallel  to  fif  and  one 
from  E  parallel  to  EF would  locate  Fas  in  Fig.  28^.  The 
polygon  is  in  the  order  of  the  arrows,  EABFE,  and  the  stresses 
in  bf  and  ef  are  consequently  127  Ibs.  compression  and  in 
Ibs.  tension  respectively.  Since  ef  and  bf  are  now  known,  the 
polygon  for  the  lower  of  the  two  joints  next  on  the  right  can 
be  completed  by  drawing  from  F  and  E  parallels  respectively 


STRESSES  IN  FRAMED  STRUCTURES. 


101 


to  fg  and  eg,  locating  G,  and  closing  the  polygon  in  which  the 
arrows  are  in  the  order  EFGE,  showing  the  stresses  vn.fg  and 
ge  to  be  173  Ibs.  compression  and  130  Ibs.  tension  respectively. 


j.  .  .  .5.°.  .  .  -1?0 

SCALE  OF  POUNDS 

FIG.  28. 

Only  two  forces  are  now  unknown  at  the  joint  where  be  is 
applied,  and  this  polygon  can  be  closed.  The  bar  fg  acted 
downward  upon  the  preceding  joint ;  it  must  therefore  act  up- 
ward at  this  joint.  GF  and  FB  of  the  polygon  are  already 


102  STATICS. 

drawn,  adding  the  known  BC,  drawing  CH  and  677  parallel 
respectively  to  ch  and  gh,  H  is  located  and  the  polygon  is 
closed  and  consists,  following  the  arrows,  of  BCHGFB,  show- 
ing stresses  in  ch  and  gh  to  be  132  Ibs.  compression  and  no 
Ibs.  tension  respectively. 

For  the  joint  at  the  lower  end  of  gh  only  eh  is  unknown, 
and  to  close  this  polygon  we  have  only  to  connect  E  and  H. 
The  arrows  are  in  the  order  EGHE,  and  eh  is  133  Ibs.  tension. 

For  the  joint  at  the  extreme  right  two  of  the  forces,  ch 
and  eh,  have  been  determined  and  cd  and  de  were  known  at 
the  outset.  It  is  important  as  a  check  to  see  if  they  are  in 
equilibrium.  Laying  off  CD  downward  and  DE  upward,  equal 
to  30  and  no  Ibs.  respectively,  E  is  found  to  fall  where  it  was 
before  located,  and  the  work  checks  with  CDEHC,  read  with 
the  arrows  as  the  last  magnitude  polygon. 

Examining  Fig.  28^,  it  appears  that  it  consists  not  only  of 
a  closed  magnitude  polygon  for  each  joint,  but  also  a  complete 
magnitude  polygon,  ABCDEA,  for  the  external  forces.  In 
fact,  the  ordinary  method  of  determining  all  the  stresses  in  a 
given  frame  under  given  fixed  loads  is,  after  equilibrium  is 
established,  to  see  that  the  external  forces  are  lettered  *  in  the 
order  of  their  occurrence  in  passing  around  the  outside  of  the 
frame,  and  complete  their  magnitude  polygon  before  construct- 
ing any  of  the  minor  ones  for  the  internal  forces  at  the  joints. 
Then  adding  two  new  lines  to  the  diagram  will  establish  the 
first  joint,  two  lines  more  the  next,  and  so  on,  each  two  lines 
completing  a  new  magnitude  polygon  until  the  last  joint  is 
reached,  where  one  bar  only  remains  to  be  determined,  and  let- 
ters are  already  in  the  diagram  which  only  need  to  be  con- 


*  It  should  be  pointed  out  that  a  letter  which  is  common  to  two  adjacent  forces 
may  be  written  once  midway  between  them  just  as  well  as  closely  adjoining  each, 
and  is  so  written  in  practice. 


STRESSES  IN  FRAMED  STRUCTURES.  103 

0 

nected  to  show  this  last  stress.  If  the  line  so  drawn  is  parallel 
to  its  corresponding  bar  the  work  checks,  and  the  results  can 
be  scaled  off  as  required.  If  the  work  does  not  so  check,  it 
shows  that  there  is  an  error  somewhere  which  must  be  found, 
and  corrected  before  the  result  for  any  bar  can  be  looked  upon 
as  trustworthy. 

76.  Stress  Diagrams. — A  diagram  such  as  just  described, 
including 

(i).  A  closed  magnitude  polygon  for  the  external  forces, 
and 

(2).  A  closed  magnitude  polygon  for  each  joint  of  the 
frame,  is  called  a  stress  diagram — sometimes  also  a  Maxwell, 
or  Cremona  diagram. 

In  constructing  stress  diagrams  as  large  a  scale  as  is  con- 
venient should  be  used  throughout,  especially  in  the  diagram 
of  the  frame,  where  otherwise  short  lines  may  give  rise  to  in- 
accuracies when  long  lines  have  to  be  drawn  parallel  to  them. 
The  equivalent  of  very  large  scale  for  frame  diagram,  with- 
out some  of  the  worst  disadvantages  of  such  large  scale,  can 
be  produced  by  drawing  in  a  group  long  parallels  to  the  bars 
through  calculated  points.  These  can  then  be  used  with  a  frame 
diagram  of  moderate  scale  or  even  a  mere  sketch  for  a  guide. 

If  a  number  of  stress  diagrams  are  to  be  drawn  from  one 
frame  diagram,  natures  of  stresses  from  different  loadings  may 
well  be  recorded  on  small  freehand  diagrams  accompanying 
each  stress  diagram. 

To  avoid  confusion,  the  lines  of  action  of  the  external  forces 
should  be  shown  entirely  outside  the  periphery  frame  as  in  Fig. 
28a. 

Each  line  of  the  stress  diagram,  except  those  representing 
the  external  forces,  represents  the  two  equal  and  opposite 
forces  which  are  in  action  at  each  end  of  the  bar  bearing  the 
same  letters  as  the  line. 


104  STATICS. 

An  error  outside  of  common  blunders  in  tne  use  of  draught- 
ing instruments,  which  is  frequently  the  cause  of  failure  of  stress 
diagrams  to  close,  is  an  incorrect  determination  of  the  reactions 
which  gives  rise  to  a  set  of  external  forces  amounting  to  a 
couple  instead  of  being  in  equilibrium,  and  such  an  error  is  not 
exposed  by  the  magnitude  polygon  for  these  forces.  The 
reactions  should  therefore  be  figured  out  independently,  i.e., 
the  correctness  of  one  should  not  be  allowed  to  depend  upon 
the  correctness  of  the  other. 

77.  General  Instructions  Regarding  Exercises  Involving 
Stress  Diagrams. — Determine  reactions,   whenever  possible, 
by  inspection,  otherwise  algebraically. 

After  completion  of  the  stress  diagram  see  that  the  work 
checks  before  proceeding  further. 

Show  upon  the  diagram  of  truss,  by  means  of  algebraic 
signs,  the  nature  of  stress  in  each  bar,  using  sign  -f-  for  com- 
pression and  —  for  tension. 

As  a  guide  in  selecting  a  place  for  magnitude  polygon  of 
the  external  forces  with  a  view  to  preventing  work  from  run- 
ning off  the  paper,  it  will  be  well  to  compute  in  advance  one 
or  two  of  the  largest  stresses,  where  the  bars  in  which  they 
will  exist  are  easily  discernible,  as  in  Exercises  21-23. 

78.  Special  Instructions  Regarding  Exercises  21-23. — 
State  for  comparison  the  graphic,  algebraic,  and  semi-algebraic 
results  for  the  bar  specified,  having  used  for  the  algebraic  work 
the  method  of  sections,  and  understanding  by  semi-algebraic 
work  a  similar  algebraic  solution  in  which  lever  arms  are  scaled 
from  the  drawing  instead  of  being  computed. 

Note  that  checking  semi-algebraically  does  not  check  the 
accuracy  of  the  frame  diagram. 

Exercise  21.  Truss  shown  in  Fig.  29.  Span  64  ft.  Rise  one  fourth 
of  the  span.  Eight  equal  panels;  1000  Ibs.  vertical  load  at  each  of  the 
joints  o  and  8,  and  2000  Ibs.  at  each  of  the  joints  1-7,  inclusive.  Scale 


STRESSES  IN  FRAMED  STRUCTURES. 


as  large  as  i  in.  =  8  ft. 
and  semi-  algebraically. 


Check  the  results  for  the  bar  2-3  algebraically 


11        12        13 

FIG.  29. 


15 


Exercise  22.  Same  truss  and  loads  as  in  Exercise  16,  but  with  addi- 
tional loads  of  4000  Ibs.  at  joints  n  and  13,  and  2000  Ibs.  at  joints  9  and 
15.  Check  the  result  for  bar  10-11  algebraically  and  semi-algebraically. 

Exercise  23.  Same  conditions  as  in  Exercise  17,  except  that  all  loads 
at  right  of  the  center  are  doubled  and  2000  Ibs.  added  at  joint  14. 

Check  the  result  for  bar  3-12  algebraically  and  semi-algebraically. 


jj 

CHAPTER   XIII. 

ADDITIONAL  TOPICS  AND   EXAMPLES. 

79.  Complications  in  Connection  with  the  Analysis  of 
Frames. — The  methods  of  the  preceding  chapter  will  be  found 
easy  of  application  to  any  simple  frame  under  vertical  loads. 
But  many  cases  are  unavoidable  in  practice  where  the  con- 
ditions are  not  so  simple,  and  a  number  of  the  most  important 
and  illustrative  of  them  will  be  taken  up  in  this  chapter.  The 
sources  of  difficulties  are  various.  Those  treated  in  the  follow- 
ing pages  fall  into  four  classes,  more  than  one  of  which  may, 
of  course,  be  found  exemplified  simultaneously  in  connection 
with  one  structure.  These  classes  of  the  sources  of  difficulty 
are 

1.  Indeterminateness,  apparent  or  real,  as  to  the  reactions. 

2.  Indeterminateness  in  the  analysis  of  a  frame,  apparent 
but  not  real, — due  to  special  systems  of  grouping  bars. 

3 .  Apparent  but  not  real  redundancy  in  framing. 

4.  Structure  being  partly  framed,  partly  non-framed. 

The  really  indeterminate  cases  here  taken  up  are  such  only 
as  can  be  brought  within  the  scope  of  statical  methods  by  mak- 
ing assumptions  of  such  reasonableness  and  of  such  general 
acceptability  in  practice 'that  the  fact  that  the  cases  are  statically 
indeterminate  is  usually  ignored. 

Structures  generally  recognized  as  redundant  do  not  fall 
within  the  scope  of  this  book. 

106 


ADDITIONAL    TOPICS  AND  EXAMPLES.  107 

0 

80.  Reactions  Due  to  Non-Vertical  Forces. — Indeter- 
minateness  as  to  the  reactions  occurs  in  the  case  of  structures 
subject  to  sidewise  or  any  non-vertical  forces,  such  as  roof- 
trusses,  towers,  etc.,  exposed  to  the  wind.  These  reactions 
are,  in  general,  two  forces  of  which  only  the  points  of  applica- 
tion are  known,  hence  involving  four  elements  unknown  (two 
slopes  and  two  magnitudes),  and  the  problem  of  finding  them 
is  indeterminate.  Results  satisfactory  in  practice  can,  how- 
ever, be  obtained  in  any  one  of  these  three  ways : 

(1)  By  supporting  one  end  upon  rollers,  assuming  them  fric- 
tionless,  thus  making  one  of  the  slopes  known. 

Rollers  are  frequently  present  anyway  in  large  roofs  to  pro- 
vide for  expansion  and  contraction,  and  such  roofs  at  once 
come  under  this  method. 

(2)  By  assuming  that  each  wall  resists  half  the  horizontal 
component  of  forces,  thus  indirectly  assuming  two  slopes. 

(3)  By  assuming  that  both  reactions  will  be  parallel  to  the 
resultant  of  all  non-vertical  loads. 

In  (i)  we  will  call  to  our  aid  a  mechanical  contrivance  and 
make  an  assumption,  and  in  (2)  and  (3)  we  make  assumptions 
only. 

Each  of  these  ways,  experience  has  justified  as  of  sufficient 
correctness. 

This  kind  of  indeterminateness  arises  whenever  the  resultant 
of  the  loads  has  a  component  parallel  to  the  supporting  surface 
or,  if  the  supports  are  in  different  planes,  to  one  or  both  of  the 
supporting  surfaces.  For  example,  in  the  case  of  a  door  or 
gate  hinged  to  a  vertical  jamb,  the  door  may  be  in  equilibrium 
however  its  weight  be  divided  between  the  hinges.  In  such  a 
case,  of  course,  if  analysis  were  necessary,  the  assumption 
would  be  that  either  hinge  may  have  to  furnish  the  whole  ver- 
tical support,  or  else  assume  it  all  on  one  or  the  other  of  the 


io8  STATICS. 

hinges,    and   by   setting   this   hinge   a   little   high,    make   the 
assumption  a  certainty. 

81.  The   Fink  Truss. — The  form   of  roof- truss  shown  in 
Fig.  30,  and  known  in  this  country  as  the  Fink  truss,  has  some 


practical  merits  which  cause  it  to  be  widely  used.  Though 
statically  determinate  it  cannot  be  analyzed  by  the  stress- 
diagram  method  without  some  special  manipulation.  On  reach- 
ing either  of  the  joints  M  or  N,  the  three  bars  P,  Q,  and  R, 
or  R,  S,  and  7",  respectively  are  unknown,  and  the  stress- 
diagram  appears  to  be  blocked.  If  any  one  of  the  five  forces 
P,  Q,  R,  5,  or  T  can  be  evaluated  the  difficulty  will  be  over- 
come. This  suggests  the  method  of  sections,  and  it  appears 
that  a  section  can  be  taken  through  T  either  side  of  the  ridge, 
which  will  cut  only  two  other  bars.  T  is  determined  accord- 
ingly. 

In  general  in  dealing  with  Fink  trusses,  the  value  of  this 
stress  might  well  be  looked  upon  as  something  always  to  be 
computed  in  advance,  as  the  reactions  are,  and  to  be  laid  off 
in  place  as  soon  as  the  magnitude  diagram  for  the  external 
forces  is  completed.  Cf.  Exercise  24. 

Exercise  24.  A  Fink  truss,  which  in  this  case  will  be  made  up  of  hor- 
izontal bars  and  bars  inclined  30  degrees  and  60  degrees  to  the  horizon, 
as  shown.  Eight  equal  divisions  in  the  upper  chord.  Load  at  each 
upper  chord  joint  2000  Ibs.  Construct  the  stress  diagram. 

82.  Triangular  Frame    with  Trussed    Top    Chord.-— A 

frame   somewhat  similar  to  the   Fink  truss  is  that  shown  in 


ADDITIONAL    TOPICS  AND  EXAMPLES^  109 

Fig.  31.  It  is  a  construction  developed  from  the  ordinary 
triangular  frame  by  a  simple  system  of 
trussing  applied  to  the  rafters.  The  re- 
sult is,  however,  that  there  is  no  joint  at 
which  a  stress-diagram  can  be  started 
immediately.  The  way  out  of  the  diffi- 
culty is,  as  in  the  case  of  the  Fink  truss, 
the  previous  determination  of  the  stress 
in  the  horizontal  tie  by  the  method  of  sections. 

The  Fink  truss  may,  in  fact,  be  regarded  as  a  special  case 
of  this  construction,  in  which  the  trussing  of  the  top  chord  is 
more  elaborate,  and  in  which  the  horizontal  tie  and  the  lower 
chord  of  that  trussing  are  coincident  to  some  extent.  Cf. 
Exercise  25. 

Exercise  25.  A  triangular  frame  of  24  ft.  span  and  20  ft.  rise  is  made 
with  its  top  chord  trussed  as  shown  (Fig.  32),  by  means  of  struts,  nor- 
mal to  them,  3  ft.  6  in.  in  length.  The  frame  is 
supported  by  a  hinge  at  the  left  end  of  the  hori- 
zontal side  and  a  set  of  rollers  at  the  right  end 
of  the  same  bar.  Loads  Pi,  P^  P>,  Pt,  and  Pi, 
are  applied  at  the  joints  and  with  the  directions 
shown.  Taking  these  forces  at  2000,  10,000, 
4000,  3000,  and  4000  Ibs.  respectively  construct 
the  stress  diagram. 

Suggestion.     Here  we  can  treat  all  external 
—  forces  as  acting   outside  the  frame,  by  dotting 
FIG.  32.  such  lines  of  action  as  may  be  needed  and  treat- 

ing such  parts  of  these  lines  as  lie  inside  the  frame  periphery  as  if  they 
were  actual  bars  of  the  frame. 

83.  Counters.— Frames  if  made  of  a  simple  triangulation 
may  be  called  upon  to  resist,  in  some  of  their  members,  both 
tension  and  compression  according  to  variations  in  loading  con- 
stantly occurring.  Such  reversal  of  stresses  is  to  be  avoided 
in  the  interests  of  economy  and  simplicity  of  design.  If  a 
diagonal  in  a  given  quadrilateral  is  replaced  by  its  mate,  if  the 
loads  remain  unchanged,  the  stresses  produced  in  the  two 


no  STATICS. 

diagonals  will  be  opposite  in  character.  Hence,  if  in  a  quad- 
rilateral where  a  diagonal  might  be  called  upon  to  resist  com- 
pression, a  mate  to  it,  if  present,  would  resist  the  stress  in 
tension,  and  no  compression  could  exist  in  either  diagonal.  A 
diagonal  added  for  such  a  purpose  is  called  a  counter. 

Counters  in  bridge  trusses  are  idle  when  the  truss  carries  a 
full  symmetrical  load.  They  are  in  action  only  under  certain 
partial  or  unsymmetrical  loads.  Reversal  of  stress  may  be 
unpreventable  in  some  members  of  certain  kinds  of  trusses. 

The  greater  the  permanent  load  compared  with  the  live, 
the  less  the  likelihood  of  reversals,  and  the  less  there  is  for 
counters  to  do. 

The  presence  of  counters  gives  the  frame  an  appearance  of 
redundancy,  but  if  the  counters  are  of  such  a  character  or 
secured  at  the  joints  in  such  a  way  that  they  are  incapable  of 
resisting  both  tension  and  compression,  the  bars  which  must 
inevitably  be  out  of  action  under  the  given  loads  can  simply  be 
ignored,  and  the  analysis  can  proceed  as  usual. 

Cf.  Exercise  26  and  Fig.  33. 

Exercise  26.  A  four-sided  framed  tower  40  ft. 
high  and  14  ft.  wide  at  the  base,  and  whose  other 
dimensions  are  as  shown  (Fig.  33),  is  made  with  a 
double  set  of  slender  diagonals  throughout.  In  a 
gale  each  half  of  it  is  supposed  to  be  subject  to  the 
forces  shown.  Assuming  half  the  horizontal  thrust 
taken  up  at  each  column  base,  construct  the  stress 
diagram. 

y  Suggestion.  In  drawing  the  diagram  of  the  frame 

dot  the  set  of  diagonals  which  are  out  of  action. 

FIG.  33-  Cf.  §  83. 

84.  Bent  of  a  Mill  Building. — A  common  and  important 
type  of  structure  subject  to  loads  elsewhere  than  at  the  joints 
and  hence  exemplifying  the  partly  framed,  partly  non-framed 
type  of  structure  is  the  combination  of  columns  with  a  truss 
constituting  the  bents  of  a  mill-building  and  shown  in  Fig. 


ADDITIONAL    TOPICS  AND  EXAMPLES. 


ill 


34.  They  are  usually  of  steel  throughout.  The  forces  which 
give  rise  to  their  peculiarities  are  sidewise  forces  due  to  wind, 
tension  of  belts,  shocks  from  traveling  cranes,  etc.  The 
general  relation  between  such  structures  and  true  frames  is 
explained  in  §64  (g.v.),  but  it  will  be  worth  while  to  amplify 
that  explanation  by  taking  as  an  example  the  bent  of  Fig.  34 


V2 

FIG.  34. 
and  discussing  in  detail  the  peculiar  steps  in  its  statical  analysis. 

The  steps  peculiar  to  the  statical  analysis  of  a  mill -building 
bent  are: 

(a)  The  assumption  of  the  points  of  application  of  the  foun- 
dation reactions,  Rl  and  Rr 

(&)  The  analysis  (by  inspection)  of  the  bent  into  its  con- 
stituent true  frame  and  flexural  members. 

(c)  The  determination  of  the  forces  brought  to  bear  upon 
the  joints  of  the  frame  by  the  flexural  members. 

The  step  (a)  is  one  made  necessary  by  the  practical  con- 
sideration that  the  columns  are  members  of  considerable 
breadth  in  the  plane  of  the  bent  even  at  their  bases.  More- 
over, they  have  to  be  anchored  to  the  foundations  so  as  to  resist 
horizontal  displacement  as  well  as  an  actual  lift.  The  result 
is  an  anchorage  which  offers  considerable  resistance  to  the  rota- 


H2  STATICS. 

tion  of  the  column  about  its  base.  This  resistance  is  of  course  a 
couple  in  the  plane  of  the  bent  which  combined  with  the  requi- 
site horizontal  and  vertical  resistances  amounts  to  the  single 
forces  R^  and  R2  intersecting  the  axes  of  the  columns  at  points 
at  some  distance  y  above  their  bases.  The  evaluation  of  j/, 
which  is  dependent  upon  the  evaluation  of  the  couples,  is  a 
statically  indeterminate  problem.  Consideration  of  the  elas- 
ticity of  the  material  *  would  lead  to  the  assumption  that  y  is 
half  the  distance  between  the  column  bases  and  the  attach- 
ment of  the  knee-braces — the  name  given  to  the  lowest  of  the 
bars  secured  directly  to  the  column.  This  would  minimize  the 
stress  in  the  columns,  but  would  necessitate  careful  attention 
to  the  design  and  execution  of  the  anchorages  and  foundations. 
In  small  structures  in  which  such  attention  to  the  anchorage  of 
the  columns  is  not  considered  worth  while,  the  existence  of  the 
couple  may  be  ignored  outright,  and  the  structure  treated  as  if 
y  were  zero  and  the  bases  of  the  columns  hinged  to  the  foun- 
dations. Then  the  couple  actually  materializing  in  the  life  of 
the  structure  is  simply  so  much  addition  to  the  factor  of  safety. 
Cases  might  arise  in  which  the  designer  would  feel  justified  in 
assuming  other  values  of  y  and  proportioning  the  anchorages 
and  other  parts  of  the  structure  accordingly. 

The  points  of  application  once  decided  upon,  the  distribution 
of  the  horizontal  components  between  the  two  reactions  (§80) 
would  be  made,  the  vertical  components  calculated  (Case  2b) 
accordingly,  and  Rl  and  Rz  established. 

Step  (£)  is  not  difficult,  for  structures  of  the  kind  in  question 
are  simple  triangulations  with  one  or  more  bars  subject  to  loads 
between  joints,  or  extended  beyond  joints  so  as  to  receive  loads 
on  the  extended  part,  usually  at  its  end.  The  simple  triangu- 
lation  constitutes  the  true  frame,  and  the  bars  just  described 

*  Cf.  Johnson,  Bryan,  and  Turneaure's  Modern  Framed  Structures,  Art.  151. 


ADDITIONAL    TOPICS  AND  EXAMPLES.  113 

are  the  flexural  members, — the  bars  spoken  of  in  §64  as  doing 
double  duty. 

Step  (c)  consists  of  finding  the  reactions  from  the  frame 
upon  the  flexural  members  necessary  to  hold  them  in  place  in 
opposition  to  the  transverse  loads  to  which  they  are  subject. 
Reversing  these  reactions  in  sense,  they  become  the  forces 
transmitted  to  the  joints  of  the  frame  by  the  flexural  members. 

The  analysis  of  the  frame  then  goes  on  as  usual,  establish- 
ing the  tension  and  compression  in  all  the  members,  including 
such  as  are  the  parts  of  the  flexural  members  common  to  the 
frame.  The  flexural  members  have  also  shears  and  flexures, 
arising  from  their  beam-action,  to  be  determined,  and  when 
this  is  done  the  statical  analysis  of  the  structure  is  complete. 

The  whole  process  is  one  proceeding  on  a  number  of 
uncertainties,  but  there  is  no  reason  why  it  may  not  surely  be 
kept  on  the  side  of  safety  and  that,  too,  without  serious  lack  of 
economy. 

The  bent  of  Fig.  34  is  accordingly  worked  out  as  follows. 

Assuming  the  columns  hinged  at  their  bases  (step  a),  and 
the  horizontal  components  Hl  and  H2  of  the  foundation  re- 
actions" to  be  equal  (§80),  the  vertical  components,  Vl  and 
V^,  of  these  reactions  follow  at  once  by  Case  2b. 

In  Fig.  35  are  shown  (step  b)  the  true  frame,  including  the 
whole  of  the  triangulation  of  bars  in  the  bent,  and,  separate 
from  it,  the  four  flexural  members, — the  two  top-chord  bars 
subject  to  transverse  loads,  and  the  two  columns  from  eaves  to 
base. 

The  two  top-chord  bars  will  naturally  be  regarded  (step  c) 
as  two  centrally  loaded  beams  requiring  reactions  of  \W  from 
the  frame  joints  at  their  ends  and  hence  transmitting  that 
amount  to  those  joints.  The  two  columns  differ  from  the  top- 
chord  bars  in  the  immaterial  particular  that  their  supporting 
joints  are  both  on  the  same  side  of  all  the  loads.  V^  and  V2 


STATICS 


being  regarded  as  transmitted  directly  up  the  column  to  the 
nearest  joint  of  the  true  frame,  the  loads  for  the  windward 
column  are  P  and  Hr  and  for  the  leeward  column  H2.  The 


S4  S4 


FIG.  35. 

reactions  Sv  S2,  S3,  and  54  are  determined  by  Case  2b.  Re- 
versing them  in  sense,  they  furnish  the  only  remaining  external 
forces  in  action  pn  the  true  frame.  The  frame  can  now  be 
analyzed  as  desired  by  familiar  methods.  Cf.  Exercise  27. 


M///////?//{//////////M 

' 


-40 


FIG.  36. 

Exercise  27.  A  bent  of  a  mill  building  is  framed  and  subject  to  loads 
as  shown  in  FIG.  36.  The  truss  proper  consists  of  four  panels  of  equal 
length.  The  columns  are  to  be  assumed  hinged  at  the  base  and  the 
horizontal  thrust  divided  equally  between  them.  (Cf.  §§  80,  84.) 


ADDITIONAL   TOPICS  AND  EXAMPLES.  "5 

9 

Construct  the  stress  diagram  and,  preferably  working  with  original 
frame  as  above  shown,  check  algebraically  the  stresses  in  Q  and  R  re- 
spectively, and  record  them  side  by  side  with  the  corresponding  graphi- 
cal results  for  comparison. 

Suggestions.  Determine  the  JFf's  and  F's  algebraically.  The  deter- 
mination of  R  will  require  the  previous  determination  of  one  other  bar. 
Show  the  bent  completely  analyzed  into  its  constituent  parts,  giving  the 
numerical  magnitudes  of  all  the  external  forces.  (Cf.  Fig.  35.) 

The  2coo-lb.  force  may  be  assumed  to  be  normal  to  the  top  chord  and 
applied  midway  between  the  two  nearest  joints.  The  knee-braces  of  this 
bent  are  seen  to  be  horizontal. 

85.  Cantilever  Bridge. — A  cantilever  bridge  consists  of 
one  or  more  trusses  or  girders  supported  at  one  or  both  ends 
by  ends  of  other  trusses  or  girders  which  overhang  their  sup- 
ports. The  overhanging  part  is  the  cantilever  whose  promi- 
nence in  this  style  of  bridge  gives  it  its  name. 

A  cantilever  bridge  always  requires  more  than  two  points 
of  support,  and  the  loads  and  reactions  constituting  usually  a 
set  of  parallel  forces,  the  reactions  would  seem  at  first  glance 
to  be  indeterminate.  Noticing,  however,  that  the  structure  is 
composed  of  at  least  two  or  three  structures,  the  reactions 
prove  to  be  determinate.  Cf.  Exercise  28. 

Exercise  28.  A  cantilever  bridge  proportioned  and  loaded  as  shown 
is  supported  by  vertical  reactions  at  A,  B,  C,  and  D.  Determine  these 
reactions  graphically  only. 

Suggestions.    Observe  that  the  hinges  at  E  and  F  divide  the  bridge 


FIG.  37- 

into  three  separate  bodies,  one  resting  upon  two  others,  that  none  of 
these  three  bodies  is  in  a  statically  indeterminate  condition,  and  that 
therefore  the  whole  structure  is  determinate. 

.Letter  as  usual  around  the  figure  and  construct  the  string  polygon 
f  jr  the   given   forces.     One   of  the  three  strings  then  missing  can  be 


n  STATICS. 

located  from  the  knowledge  that  the  flexure  at  E  and  F  must  be  zero 
(§  70),  and' the  other  two  follow  at  once. 

86.  Three-hinged  Arch. — The  structure  typified  in  Fig.  38, 
consisting  of  two  ribs  (framed  or  non-framed,  straight  or 
curved,  but  usually  curved)  which  are  hinged  together  at  their 
upper  ends  C  and  rest  on  hinges  at  their  lower  ends  A  and 

c 


FIG.  38. 

Bj  is  called  a  three-hinged  arch.     It  is  used  for  roofs  where  the 
greatest  spans  are  required,  and  for  bridges  as  well. 

It  is  a  structure  which  evidently  cannot  stand  up  even  under 
vertical  loads  without  horizontal  resistances  at  the  supports. 
In  this  respect  it  is  like  an  arch,  and  it  easily  lends  itself  to  the 
pleasing  curved  form  of  an  arch  if  desired.  Moreover,  it  is  the 
only  statically  determinate  method  of  arch  construction,  and  an 
important  and  fruitful  subject  of  study  accordingly. 

There  appears  to  be  an  indetermination  in  connection  with 
the  reactions  at  A  and  B.  Here  two  forces  are  required  of 
which  only  the  points  of  application  are  given.  Two  magni- 
tudes and  two  slopes  must  be  determined,  or  else,  if  each 
reaction  be  replaced  by  two  convenient  components,  four 
magnitudes. 

Now,  the  hinge  at  C  (assumed  like  the  other  hinges  to  be 
frictionless)  cannot  resist  any  force  which  does  not  pass  through 
it,  and  this  fact  in  connection  with  the  three  general  conditions 
of  equilibrium  furnishes  the  basis  for  the  necessary  four  equa- 
tions and  the  problem  is  seen  to  be  determinate.- 

These  four  conditions  would  be  satisfied  if  two  forces  were 


ADDITIONAL    TOPICS  AND  EXAMPLES.  117 

m> 

determined  which,  if  applied  at  A  and  B  respectively,  would 
combine  with  all  the  forces  between  those  points  and  C  and 
yield  two  equal,  opposite,  and  coincident  resultants  passing 
through  C.  These  two  forces  would  be  the  required  reactions. 
Their  intersection  in  the  magnitude  polygon  would  be  the  pole 
which  (§  54)  would  direct  a  string  polygon  for  the  given  loads 
through  the  three  given  points  A,  B>  and  C  in  such  a  way 
that  the  strings  through  A  and  B  would  be  the  extreme  strings 
of  the  whole  given  set,  and  the  one  through  C  the  string  com- 
mon to  the  resultants  of  the  two  groups  on  each  side  of  C. 

Conversely,  such  a  pole  once  determined,  the  two  lines 
from  it  to  the  ends  of  the  magnitude  polygon  for  the  given 
forces  would  give  the  magnitudes  and  directions  required. 

The  methods  of  §  54  will  therefore  furnish  the  graphic  and 
algebraic  methods  for  the  determination  of  the  reactions. 

Moreover,  the  string  polygon  through  these  three  points  for 
forces  lettered  in  the  order  of  their  occurrence  is  the  locus  of 
the  intersection  (with  any  section)  of  the  resultant  of  all  the  ex- 
ternal forces  on  either  side  of  the  section.  What  is  more, 
each  string  is  the  resultant  of  all  the  external  forces  on  either 
side  of  it  and  its  magnitude  and  sense  can  be  found  from  the 
corresponding  ray. 

This  string  polygon  once  drawn,  the  stress  in  any  bar  under 
the  most  complicated  system  of  loading  can  be  determined  by 
the  method  of  sections  (§  73)  with  an  equation  of  moments 
involving  only  four  quantities,  the  known  magnitude  of  the  re- 
sultant, the  required  magnitude  and  their  respective  lever  arms. 
These  last  can,  of  course,  be  scaled  or  calculated,  but  as  a 
careful  drawing  is  necessary  in  any  event  for  the  construction 
of  the  string  polygon,  scaling  would  commonly  be  far  more 
convenient. 

An  alternative  solution  of  the  three-hinged  arch  is  some- 
times given  which  is  based  upon  the  fact  that  it  may  be 


n8  STATICS. 

regarded  as  two  separate  structures,  and  the  groups  of  loads 
coming  on  each  assumed  to  act  separately.  Thus  (Fig.  38)  if 
the  portion  AC  be  unloaded  while  CB  is  loaded,  the  direction 
of  the  reaction  at  A  must  be  through  A  C,  and  this  reaction  on 
CB  and  the  one  at  B  are  determinate,  falling  under  Case  3. 
Repeating  the  process  for  AC,  two  more  partial  reactions  are 
determined.  The  resultants  of  the  two  reactions  at  each 
hinge  are  the  reactions  required  for  the  structure. 

It  should  be  observed  that  the  hinges  A  and  B  may  be  put 
at  different  levels  ^without  affecting  the  determinateness  of  the 
problem  or  its  method  of  treatment. 

The  horizontal  components  at  A  and  B  may  be  furnished 
by  abutments  as  with  any  arch,  or,  in  cases  where  circum- 
stances permit,  by  tie-rods  connecting  A  and  B,  together 
with  such  anchorage  as  would  be  required  with  the  given 
load  to  keep  any  truss  from  moving  as  a  whole.  With  the 
tie-rod  the  structure  becomes  what  might  be  regarded  as  a 
triangular  truss,  two  of  whose  members  are  more  or  less  curved 
and  more  or  less  frameworks  themselves. 

87.  Line  of  Pressure. — The  line  of  pressure  in  any  struc- 
ture is  the  locus  of  the  intersections  with  successive  section 
planes  throughout  the  structure  of  the  resultants  of  all  the 
external  forces  on  either  side  of  those  sections. 

In  case  the  loads  are  non-continuous,  the  line  of  pressure 
will  be  a  broken  line;  with  continuous  loads,  it  is  a  curve 
which  must  be  plotted  point  by  point  at  sections  taken  at  short 
intervals. 

An  example  of  the  former  case  arises  in  connection  with 
the  three-hinged  arch  under  a  series  of  concentrated  loads. 
The  string  polygon  including  the  end  reactions  as  extreme 
strings  and  passing  through  the  three  hinges  is  the  line  of  pres- 
sure of  that  arch  for  the  given  loads,  if  the  loads  be  lettered 
in  the  order  of  their  occurrence. 


ADDITIONAL    TOPICS  AND  EXAMPLES. 


119 


An  example  of  the  latter  case  is  a  masonry  dam  exposed 
as  it  is  to  uninterrupted  hydrostatic  pressure.  Points  in  the 
line  of  pressure  are  located  in  this  case  by  taking  successive 
horizontal  sections  and  finding  the  resultants  of  the  water 
pressure  and  weight  of  the  masonry  above  that  section.  Where 
these  resultants  pierce  the  sections  which  define  and  limit  their 
components  are  points  in  the  line  of  pressure  of  the  section  of 
the  dam. 

Exercise  29.     A  three-hinged  arch  of  240  ft.  span  is  subject  to  loads 
as  shown  in  Fig.  39.   The  top  chord  is  divided  into  eight  equal  bays,  and 


3000 


FIG.  39. 

the  bays  of  the  bottom  chord  are  each  equal  to  AE  in  length,  DE  being 
horizontal.     The  joints  of  the  bottom  chords  are  on  the  arc  of  a  circle 
of  145  ft.  radius  through  A  and  B.     Other  dimensions  as  shown. 
Required  : 

1.  Reactions  graphically  by  the  method  of  §  54. 

2.  Reactions  algebraically  (§  54),  showing  all  lever  arms  dimen- 

sioned except  such  as  are  given  directly  by  the  main  dimen- 
sions. 

3.  Line  of  pressure. 

4.  Stress  diagram. 

Note  that  the  stress  diagrams  of  frames  A  C  and  BC  will 
each  have  its  own  check. 

5.  Pressure  on  the  hinge  at  C  graphically. 

6.  Stress  in  the  bars  marked  Q,  R,  and  S  with  the  aid  of  the  line 

of  pressure  and  scaled  arms. 

7.  Stress  in  Q,  R,  and  S  from  the  stress  diagram. 

8.  Tabulated  record  of  results  of  i,  2,  5,  6,  and  7. 


120 


STATICS. 


Suggestions  .  Use  a  large  scale  for  the  frame  diagram,  say  i  in.  =  30 
or  40  ft.,  and  a  moderate  scale  for  the  magnitude  diagram. 
Is  the  line  of  pressure  a  closed  string  polygon  ? 

Exercise  30.  Suppose  the  coordinates  of  the  three  hinges,  A,  B, 
and  C,  of  a  three-hinged  arch  to  be  expressed  in  feet  as  (o,  o),  (60,  30), 
and  (30,  50)  respectively,  and  the  ribs  to  be  of  any  curvature  and  either 
framed  or  non-framed.  Suppose  a  vertical  load  of  20  tons  acting  20  ft. 
horizontally  from  A,  and  a  horizontal  force  of  10  tons  acting  towards 
the  right  and  applied  between  B  and  C  and  10  ft.  vertically  above  B. 
Determine  the  reactions  algebraically. 

Suggestion.  The  horizontal  and  vertical  components  of  the  two 
reactions  would  usually  be  a  more  convenient  form  for  the  expression 
of  the  required  answer  than  the  resultant  reactions  themselves. 

Use  the  method  of  §  54,  and  check  the  results  carefully. 

88.  Hammer-beam   Truss. — Like   all  other   roof  trusses, 
the  hammer-beam  truss,  Fig.  40,  has  vertical  loads  to  resist 


FIG.  40. 

and  also  non-vertical  ones.  Unlike  most  trusses,  some  of  its 
main  members  are  curved.  This  curvature  alone  would  not 
call  for  any  special  treatment  here,  for  curved  members  may, 
for  purposes  of  analysis,  be  considered  replaced  by  straight 
ones  connecting  the  same  joints  and  the  curvature  left  out  of 
account  till  the  design  of  the  piece  itself  is  taken  up.  The 
truss  would  then  be  a  complete  frame  and  would  be  treated 
like  any  other. 

The  combined  effect  of  two  circumstances,  however,  ren- 


ADDITIONAL    TOPICS  AND  EXAMPLES.  «* 

• 

ders  it  necessary  to  approach  the  hammer-beam  truss  in  a  dif- 
ferent way.  These  two  circumstances  are,  (i)  curved  bars  in 
frames  at  best  work  under  very  unfavorable  conditions,  arid  (2) 
in  these  trusses  they  are  made  of  wood, — a  material  specially 
weak  and  difficult  to  connect.  Moreover,  even  at  best  the 
shape  of  this  truss  is  in  itself  unfavorable  to  rigidity.  The 
result  is  that  even  under  perfectly  vertical  loads  the  hammer- 
beam  truss  must  be  expected  to  produce  thrust  upon  each  sup- 
porting wall,  as  an  arch  would  do,  or  as  rafters  without  any 
collar-beam  (or  with  the  collar-beam  too  close  to  the  ridge) 
would  do.  That  means  that  the  walls  must  be  prepared  to 
resist  the  spreading  at  the  base  of  the  truss  which  the  curved 
ties  are  unable  to  prevent. 

To  estimate  this  thrust  from  vertical  loads,  a  good  way  is 
to  place  no  reliance  whatever  upon  the  upper  curved  members 
— to  consider  them  absent.  This  reduces  the  truss  to  an  in- 
complete frame — the  parts  below  the  joints  3  and  7  are  mere 
framed  inclined  struts,  reaching  from  their  foothold  on  the 
wall  to  joints  3  and  7  of  the  truss  3,  4,  5,  6,  7,  12  which  they 
support.  The  loads  on  this  truss  are  known  and  hence  the 
vertical  load  at  the  top  of  each  strut.  The  vertical  reaction  at 
the  base  of  each  strut  is,  of  course,  equal  and  opposite,  but  as 
it  is  not  coincident  with  it,  the  two  form  a  couple,  and  the 
struts  would  fall  into  the  building  but  for  horizontal  resistances 
at  their  upper  ends  along  the  bar  3,  12,  7.  This  horizontal 
resistance  calls  for  an  equal  and  opposite  one  at  the  base  of  each 
strut.  These  horizontal  forces  constitute  the  second  couple 
which  balances  the  first.  The  common  magnitude  of  the  hori- 
zontal forces  may  be  taken  as  the  thrust  of  the  roof,  and  with 
the  vertical  reactions,  which  are  easily  found,,  there  is  no 
further  obstacle  to  the  construction  of  the  stress  diagram. 

Finding  thrusts  as  just  described  and  making  no  claim  on 
the  upper  curved  members,  and  not  an  unduly  severe  one  on 


122  STATICS. 

lower  ones,  leaves  the  former  ready  to  act  at  their  full  value 
when  the  wind  blows,  and  the  latter  in  as  favorable  a  condition 
as  possible. 

In  determining  wind  stresses,  the  only  thing  to  be  done 
(except  as  described  below)  is  to  treat  the  truss  as  a  complete 
frame,  as  explained  above.  The  results  thus  obtained  will  be 
combined  with  those  from  the  vertical  loads. 

The  only  difficulty,  then,  is  to  provide  for  the  stresses.  The 
stiffness  of  the  rafters  at  7  may  have  to  be  relied  upon  to  assist 
the  upper  curved  members,  and  the  buttresses  may  have  to  rise 
to  the  level  of  I  and  9  (as  shown  in  the  figure)  in  order  to 
relieve  the  lower  curved  members. 

This  last  would  raise  real  points  of  support  of  the  truss  to  I 
and  9,  and  9,  10  and  o,  I  would  become  mere  posts  prevented 
from  overturning  by  the  buttresses,  and  o,  1 1  and  10,  13  mere 
ornaments. 

Of  course  in  a  large  roof  it  may  be  necessary  to  rely  upon 
the  stiffness  of  the  rafters,  and  even  then  higher  stresses  than 
would  ordinarily  be  regarded  as  satisfactory  may  have  to  be 
tolerated.  It  may  be  advisable  even  to  neglect  all  curved 
members  and  rely  entirely  on  high  buttresses  and  the  stiffness 
of  the  rafters. 

The  hammer-beam  truss  is  a  very  imperfect  structure  as 
regards  the  economical  application  of  material.  The  less  the 
curvature  of  members  the  better  from  this  point  of  view. 

Exercise  31.  With  a  vertical  load  of  1000  Ibs.  per  bay  of  the  top 
chord  of  a  hammer-beam  truss  such  as  is  shown  in  Fig.  40,  determine  the 
thrust  on  the  walls  on  the  assumption  that  the  upper  pair  of  curved 
bars  is  wholly  inoperative,  and  construct  the  stress  diagram.  All  bars 
of  the  truss  are  to  be  taken  as  horizontal,  vertical,  or  inclined  at  45°  as 
shown,  and  o,  i  and  9,  10  each  of  the  same  length  as  i,  1 1  or  9,  13. 

Suggestion.  Draw  a  separate  diagram  of  one  of  the  framed  struts 
o,  i,  2,  3,  n,  showing  all  the  loads  upon  it,  and  determine  the  reactions 
upon  it  as  an  independent  structure.  The  way  is  then  clear  for  the 
stress  diagram  as  usual. 


ADDITIONAL    TOPICS  AND  EXAMPLES,  123 

89.  Stresses  Due  to  Moving  Loads. — Some  structures, 
as  bridges  and  viaducts,  are  subject  to  loads  varying  greatly 
in  magnitude  and  taking  widely  varying  positions  on  the 
structure. 

The  special  steps  involved  in  the  determination  of  the 
stresses  in  such  a  structure  are 

(1)  The  assumption  of  the  maximum  reasonable  load  or 
system  of  loads. 

(2)  The  determination  of  the  locations  of  this  load  or  sys- 
tem of  loads  which  will  produce  the  extremes  of  stress  in  each 
and  every  part  of  the  structure. 

(3)  The  determination  successively  of  the  extremes  of  stress 
in  each  and  every  part  of  the  structure  due  to  loads  assumed 
to  be  stationed  at  all  the  various  points  decided  by  the  preced- 
ing step. 

The  first  of  these  steps  is  an  exercise  of  judgment  in  the 
light  of  experience  of  what  the  structure  may  reasonably  be 
expected  to  carry,  leaning  of  course  towards  too  high  rather 
than  too  low  an  estimate. 

The  second  of  the  steps  opens  a  large  field  in  the  study  of 
special  structures  and  systems  of  loading.  In  general  it  may 
be  said  to  be  a  process  based  upon  the  observation  of  the  stress 
in  a  given  part  of  a  structure  due  to  a  single  force  followed  by 
a  study  of  how  and  to  what  extent  other  forces  may  be  grouped 
with  it  to  reinforce  its  effect.  The  methods  are  a  simple 
development  of  the  general  principles  of  statics  as  treated  in 
Part  I  with  the  addition  of  various  devices,  largely  graphical, 
for  saving  time  and  labor. 

The  third  step  is  a  straightforward  solution  of  the  simple 
statical  problem  of  finding  the  stress  in  a  certain  part  of  any 
structure  subject  to  a  given  set  of  loads. 

The  reader  interested  further  in  this  subject  will  find  it 
treated  at  length  in  Burr's  Stresses  in  Bridge  and  Roof 


124  STATICS. 

Trusses,  Johnson,  Bryan,  and  Turneaure's  Modern  Framed 
Structures,  Merriman  and  Jacoby's  Roofs  and  Bridges, 
Parts  I  and  II,  Hoskins'  Graphic  Statics,  DuBois'  Stresses 
in  Framed  Structures,  etc. 

The  suddenness  of  the  application  of  the  moving  load  or 
impact  is  not  considered  at  all  in  this  analysis.  Allowance  is 
made  for  that  by  certain  empirical  methods  in  connection  with 
the  proportioning  of  the  parts  to  resist  the  extreme  stresses  de- 
termined as  just  described. 

90.  Stability  of  a  Masonry  Dam. — The  lines  of  pressure 
for  the  various  conditions  to  which  a  masonry  dam  is  exposed 
afford  a  very  important  aid  in  the  study  of  the  stability  and  de- 
sign of  such  a  structure. 

To  construct  such  a  line  of  pressure  the  cross-section  of  the 
dam  is  drawn  to  a  large  scale  and  divided  into  a  suitable  num- 
ber of  strips  by  a  series  of  horizontal  planes.  The  line  of 
pressure  will  be  located  when  there  have  been  found  the  inter- 
sections with  each  of  these  planes  of  the  resultant  of  all  the 
forces  acting  upon  the  dam  above  that  plane. 

The  gravity  and  hydrostatic  forces  on  each  strip  are  then 
determined  (assuming  the  thickness  of  the  strip  normal  to  the 
paper  to  be  one  foot)  and  shown  in  their  proper  places. 

This  being  done  the  successive  resultants  can  be  located  by 
Case  I.  The  forces  can  sometimes  be  most  conveniently  let- 
tered by  beginning  with  the  top  strip  and  lettering  the  gravity 
forces,  in  the  order  of  occurrence  of  their  strips,  ab,  be,  etc., 
and  lettering  downward,  giving  the  hydrostatic  pressures  in  a 
similar  order  the  letters  ab' ,  b'c' ,  etc. 

A  general  string  polygon  will  locate  the  successive  result- 
ants bb ',  cc' ,  etc.,  and  their  intersections  with  the  lower  limits 
of  the  successive  portions  of  the  profile  to  which  they  belong 
will  be  points  of  the  line  of  pressure. 

An  alternative  lettering  suitable  to  some  cases  would  be  to 


ADDITIONAL    TOPICS  AND  EXAMPLES.  125 

f 

letter  the  gravity  and  hydrostatic  forces  on  the  strips  respec- 
tively ab  and  be,  cd  and  de,  etc.,  beginning  at  the  top.  Tak- 
ing the  pole  coincident  with  A,  the  strings  oc,  oe,  etc.,  would 
be  the  required  resultants. 

Sometimes  it  may  be  more  satisfactory  to  determine  the 
locations  of  the  resultants  for  the  successive  groups  of  gravity 
forces  algebraically,  and  use  the  string  polygon  only  for  the 
hydrostatic  forces.  The  intersections  of  the  pairs  of  partial 
resultants  will  be  points  of  the  series  of  required  resultants 
which  can  then  be  shown  by  transference  from  the  magni- 
tude diagram. 

The  location  of  the  line  of  pressure  in  masonry  arches 
differs  from  the  preceding  process  only  in  the  greater  uncer- 
tainty about  the  elements  of  the  external  forces  on  the  struc- 
ture and  the  lack  of  a  determinate  point  for  beginning  the  line 
of  pressure  within  the  structure. 

Exercise  32.  A  masonry  dam  with  vertical  upstream  face  has  a  cross- 
section  determined  by  the  following  coordinates  (origin  at  the  top  of 
the  upstream  face)  :  (20,0),  (20,  -  17),  (21,  -  30),  (31,  -  50),  (43,  -  70), 
(58,  -  90),  (73,  -  1 10),  (o,  -  1 10). 

Required  the  lines  of  pressure  (a)  when  the  reservoir  is  empty,  and 
(b)  when  the  reservoir  is  full. 

Weight  of  masonry  to  be  taken  at  1 50  Ibs.  per  cu.  ft.  and  of  water  at 
62.5  Ibs. 

Statical  work  to  be  done  graphically  only. 

91.  Action  and  Reaction  not  Necessarily  Normal  to  the 
Surface  of  the  Contact  of  the  Bodies. — The  function  of 
statics  is  to  deal  with  the  equilibrium  of  bodies  without  regard 
to  the  origin  of  the  forces  acting  on  them.  In  order,  however, 
to  deal  with  questions  of  equilibrium  it  is  necessary  to  be 
familiar  with  the  various  possible  sources  of  forces  and  the 
conditions  under  which  they  are  effective  in  order  correctly  to 
prepare  a  problem  for  the  application  of  purely  statical  methods. 

In  the    exercises   prescribed  in   the  foregoing   work,    the 


126  STATICS. 

sources  of  all  the  forces  dealt  with  might  be  classified  broadly 
as  either  directly  or  indirectly  the  attraction  of  gravitation, 
which  needs  no  further  comment,  and  to  an  interaction  be- 
tween masses  most  naturally  thought  of  as  normal  to  their 
surface  of  contact.  Accordingly  special  attention  will  now 
briefly  be  given  to  the  possibility  and  causes  of  interactions 
between  masses  inclined  to  the  normal  to  their  surface  of 
contact. 

92.  Friction. — It  is  a  property  of  all  bodies  that  they 
offer  more  or  less  resistance  to  being  moved  over  one  an- 
other. This  resistance  is  a  force  in  the  plane  of  contact 
always  to  be  considered  when  one  body  rests  upon  another 
and  the  other  forces  acting  on  the  body  have  an  unbalanced 
component  parallel  to  that  plane.  This  kind  of  resistance  is 
called  friction.  The  resultant  of  the  friction  with  the  normal 
resistance  is  a  force  acting  on  the  body  inclined  to  the  normal 
to  their  surface  of  contact. 

The  elements  of  a  force  due  to  friction  have  the  following 
characteristics. 

The  point  of  application  may  always  be  taken  in  the  sur- 
face of  contact. 

The  direction  is  always  opposite  to  any  unbalanced  com- 
ponent of  the  other  forces  parallel  to  the  surface  of  contact. 

The  magnitude,  like  that  of  any  other  passive  resistance,  is 
variable.  It  is  as  large  as  it  has  to  be  (up  to  a  certain  limit)  to 
maintain  equilibrium  and  no  larger.  In  this  respect,  friction, 
or  tangential  resistance,  is  precisely  like  the  normal  resistance. 
The  limit  in  the  latter  case  is  the  ultimate  compressive  strength 
of  the  weaker  of  the  two  bodies  and,  in  the  former,  the  limit 
is  a  certain  percentage  (called  the  coefficient  of  friction)  of 
the  normal  pressure  existing  at  the  time  in  question.  If  the 
former  be  overstepped  penetration  or  crushing  will  occur,  if 
the  latter,  sliding  or  rolling.  Both  limits  are  experimentally 


ADDITIONAL    TOPICS  AND  EXAMPLES.  I*? 

determined  for  various  materials  and  conditions,  and  data  re- 
garding them  can  be  found  on  record  in  the  standard  reference 
books. 

The  coefficient  of  friction  is  commonly  designated  by  the 
letter  //.  Its  values  vary  greatly  according  to  the  materials  in 
contact,  the  smoothness  of  the  surface  of  contact,  the  degree  of 
lubrication,  whether  the  motion  is  likely  to  be  sliding  or  rolling. 

When  the  limit  of  friction  is  reached  and  the  body  is  about 
to  move,  the  reaction  upon  this  body  from  the  one  under  it  is 
at  a  limiting  inclination  to  the  normal  to  the  surface  of  con- 
tact called  the  angle  of  friction.  The  tangent  of  this  limiting 
angle  is  evidently  the  ratio  of  the  limiting  frictional  resistance 
to  the  normal  pressure  with  which  it  is  associated.  That  is, 
the  coefficient  of  friction  is  the  tangent  of  the  angle  of 
friction. 

If  greater  resistance  to  motion  on  the  surface  of  contact  be 
required  than  friction  can  be  trusted  to  furnish,  recourse  can  be 
had  to  various  devices,  prominent  among  them  the  use  of  one 
or  more  bodies  penetrating  each  of  the  two  given  ones,  pass- 
ing through  their  surface  of  contact  and  furnishing  the  required 
resistance  in  that  surface  by  virtue  of  their  shearing  strength. 
Nails,  bolts,  rivets,  and  dowels  are  examples  of  the  bodies 
used  for  such  purpose. 

What  is  of  most  importance  with  regard  to  friction  from 
the  point  of  view  of  statics  is  that  it  is  a  source  of  passive  force 
tangential  to  a  surface  of  contact  between  two  bodies  just  as 
normal  resistance  is  a  source  of  passive  force  normal  to  that 
surface.  Of  course  a  force  due  to  friction  is  treated  statically 
just  like  any  other  force. 

Example  i.  A  body  of  weight  W  rests  upon  a  rough 
plane  inclined  a  to  the  horizon  and  is  subject  to  a  horizontal 
force  P.  The  coefficient  of  friction  being  /^,  between  what 
two  limits  may  P  vary  while  the  body  remains  at  rest  ? 


128  STATICS. 

Solution.  The  forces  acting  on  the  body  are  the  active 
forces  P  and  W,  and  the  passive  ones  N  and  T—  i*N.  When 
Tacts  downward  as  shown,  Fig.  41,  P  is  at  its  upper  limiting 

value  and  the  body  is  about  to  move 
up  the  plane. 

Statically  the  problem  is  one  to 
determine  two  magnitudes  in  a  set 
of  concurrent  forces  whose  lines  of 
action  are  all  given  (Case  20).     The 
FIG.  41.  unknowns  are  P  and   N,    T  being 

known  as  soon  as  N  is.  The  regular  solution  of  Case  20. 
can  now  be  applied.  No  numerical  data  being  given  the 
algebraic  method  will  be  used. 

Resolving  the  four  forces  along  and  at  right  angles  to  the 
plane  for  the  Ax  and  Ay  equations  there  results 

P  cos  a  —  i*N  —  J^sin  a  =  O, 
P  sin  a—  N  -|-  Wcos  a  =  O. 
Eliminating  N,  it  appears  that 

sin  a  -\-  ft  cos  a 
~  cos  a  —  JA  sin  a 

Reversing  the  direction  of  T  the  other  limiting  value  of  P  is 
found  by  similar  means  to  be 

sin  OL  —   u.  cos  fx  _ 


p  = 


, 

cos  a  -|-  fj.  sin  a 

The  same  result  might  have  been  reached  by  reversing  the 
sign  of  /*  in  the  other  value  of  P. 

Example  2.  A  uniform  ladder  rests  upon  a  horizontal  floor 
and  against  a  vertical  wall.  The  floor  and  wall  are  of  differ- 
ent materials  and  the  coefficients  of  friction  between  them  and 
the  material  of  the  ladder  are  respectively  .30  and  .20.  What 
is  the  least  inclination  to  the  horizon  at  which  the  ladder  can 
rest  ? 


ADDITIONAL    TOPICS  AND  EXAMPLES. 


129 


Solution.  The  ladder  is  to  be  in  equilibrium  under  three 
forces  :  W  its  weight  acting  at  its  center 
of  gravity,  P  and  Q  at  its  upper  and 
lower  ends  respectively  with  directions 
inclined  to  the  normals  to  the  wall  and 
floor  by  their  respective  angles  of  fric- 
tion, and  in  such  a  way  as  to  give  P  an 
upward  component  and  Q  a  component 
towards  the  wall.  The  part  that  pure 
statics  plays  in  this  problem  is  to  furnish 
the  condition  that  W  must  pass  through 
the  intersection  of  P  and  Q.  To  this 
end  the  middle  of  the  ladder  must  be 
in  the  same  vertical  with  the  intersec- 
tion of  P  and  Q.  The  inclination,  0,  of  the  ladder  when  this 
condition  is  satisfied  will  be  the  limiting  inclination  required. 

The  determination  of  'this  inclination  is  henceforth  purely 
a  geometric  and  trigonometric  process.  Thus  can  be  written 
(Fig.  42), 

cos  (0  +  02)        sin  (0  +  0J 


sn 


cos  6l 


whence  can  be  established  tan  0  =  -  —  -,  and  6^  and 

6l  being  given  respectively  as  tan-1  0.30  and  tan-1  o.so,   it 
follows  that 

0  ==  tan-1  1.567  =  57°  27'.  Ans. 

Additional  problems  involving  friction  may  be  found  in 
abundance  in  such  works  as  those  mentioned  at  the  end  ol 
Chapter  VI. 


APPENDIX. 
ADDITIONAL  REGARDING  THE  SCOPE  OF  PURE  STATICS. 

IN  this  Appendix,  as  in  Chapter  V,  the  term  Pure  Statics  is 
used  to  designate  the  treatment  of  problems  in  which  the  data 
consist  exclusively  of  elements  of  forces,  the  desiderata  in- 
clude only  other  elements  of  forces,  and  the  conditions  of  equili- 
brium of  rigid  bodies  form  the  only  determining  conditions. 

A  somewhat  different  class  of  statical  problems  is  some- 
times met,  especially  in  text-books,  in  which  the  object  is  to 
determine  the  position  in  which  a  given  body  will  rest  under 
given  forces  and  with  given  conditions  of  support.  The  de- 
siderata here  are  geometrical  relations  between  bodies  con- 
sistent with  equilibrium,  and  not  elements  of  forces.  Such 
problems  might  not  improperly  be  called  geometrico-statical 
problems  to  distinguish  them  from  purely  statical  problems. 
An  illustration  of  such  is  to  be  found  in  Example  2  of  §  92 . 
Their  solution  consists  in  using  the  conditions  of  equilibrium 
to  identify  the  position  of  the  body  for  which  the  forces  acting 
would  be  interrelated  in  a  manner  consistent  with  equilibrium, 
and  then  in  using  sufficient  geometric  insight  to  derive  a  simple 
definition  of  that  position, — the  latter  of  the  two  steps  often 
proving  the  more  puzzling  of  the  two. 

Moreover,  as  stated  in  §46,  a  large  part  and  often  the 
most  difficult  part  of  an  ordinary  statical  problem  in  practice 
is  the  discernment  of  the  data  in  proper  shape  for  insertion  into 
the  purely  routine  processes  of  pure  statics. 

130 


APPENDIX.  131 

f 

Though  pure  statics  must  accordingly  be  understood  to 
have  a  restricted  meaning,  nevertheless  within  its  domain  will 
be  found  nearly  all  of  the  statical  problems  of  engineering. 
The  mathematical  extent  of  this  domain  offers  an  inviting  field 
of  inquiry  with  a  view  to  finding  out  how  many  and  what 
problems  might  arise,  which  would  not  be  obviously  beyond 
its  purview  by  involving  more  unknowns  than  there  are  deter- 
mining conditions. 

As  a  simple  matter  of  permutations  and  combinations  an 
early  limit  is  set  to  the  number  of  such  problems.  It  is  merely 
a  question  of  the  number  of  ways  in  which  groups  of  elements, 
each  group  containing  as  many  elements  as  there  are  determin- 
ing equations,  can  be  selected  without  repetition  from  the  total 
number  of  elements  which  pertain  to  forces  not  exceeding 
in  number  the  determining  equations.  It  will  be  convenient 
to  treat  forces  in  two  groups,  (a)  non-concurrent  (excluding 
parallel)  forces  and  (£)  concurrent  (including  parallel)  forces, 
in  which  the  determining  equations  are  three  and  two  re- 
spectively, and  the  limiting  number  of  combinations  eighty- 
four  and  fifteen  as  deduced  in  §44.  As  implied  in  §44,  many 
of  these  combinations  are,  after  all,  statical  identities.  For 
example,  a  magnitude,  direction,  and  point  of  application  can 
be  selected  from  a  group  containing  three  each  of  these  ele- 
ments in  three  different  ways,  but  in  each  case  the  statical 
significance  is  the  same,  viz.,  that  the  magnitude,  direction, 
and  point  of  application  of  one  of  a  group  of  forces  are  to  be 
evaluated.  Collecting  the  statical  identities  as  separate  cases, 
there  are  found  for  non-concurrent  forces  only  twenty  such 
cases,  numbered  in  the  following  table  I-XX,  and  for  concur- 
rent only  nine,  numbered  XXI-XXIX. 

It  should  be  borne  in  mind  that  for  non-concurrence,  as 
here  understood,  it  is  necessary  and  sufficient  that  the  resultant 
of  all  the  known  forces  shall  not  include  a  point  common  to 


I32  STATICS. 

three  unknowns  nor  be  parallel  to  them,  and  for  concurrence 
that  this  resultant  shall  intersect  the  point  common  to  two 
unknowns  or  be  parallel  to  them.  It  hardly  need  be  added 
that  with  non-concurrent  forces  if  three  unknowns  should  have 
a  point  in  common  or  be  parallel,  the  problem  would  always 
be  incapable  of  solution. 

The  following  table  gives  a  list  of  all  the  twenty-nine 
cases,  with  the  number  of  statical  identities  included  in  each 
case.  P,  a,  and  m  are  used  to  designate  magnitude,  direc- 
tion, and  point  of  application  respectively  of  any  of  three 
forces,  Q,  Ry  and  5,  which  may  involve  unknowns.  The 
sense  may  be  understood  to  be  unknown  whenever  either  the 
magnitude  or  the  direction  is  unknown. 

The  reader  who  has  mastered  the  four  cases  of  §45  will 
find  the  additional  ones  an  interesting  field  for  further  practice 
with  the  same  methods,  adjusting  them  of  course  to  the 
peculiarities  of  each  case.  The  graphical  method  is  recom- 
mended as  especially  convenient  for  this  investigation.  The 
problem  in  the  first  twenty  cases  is  then  uniformly  to  close  a 
magnitude  polygon  and  a  string  polygon,  observing  that  in 
the  last  nine  cases,  insuring  the  concurrence  of  the  forces  may 
replace  the  closure  of  the  string  polygon.  The  number  of 
forces  involved  need  never  exceed  four,  any  number  of  given 
forces  being  considered  to  be  represented  by  any  single  force 
as  their  resultant.  Some  of  these  new  cases  will  be  found  in- 
capable of  solution,  or  capable  of  one  or  more  real  solutions, 
according  to  the  relations  between  the  data  in  each  case. 
Others  are  always  indeterminate.  Very  few  are  always  capa- 
ble of  only  a  single  solution. 

Some  of  the  last  nine  cases  might  be  regarded  as  special 
cases  already  included  in  the  first  twenty. 


APPENDIX. 


133 


No. 

Unknown  Elements  Pertainine;  to 

No.  of 
Statical 
Identities 
Included. 

Remarks. 

Q 

R 

S 

I 

p 

a 

m 

P 

a 

m 







3 

Case  1  of  §  45 

•8 

I 

j5 

1 
§ 

6 

fc 

II 

p 

6 

Case  3  of  §  45 

III 

p 



P 

6 

IV 

a 

a 

m 

6 

V 

a. 

P 
P 

a 



6 

VI 

m 

m 

6 

VII 

m 

P 
P 

P 

a 

m 

6 

VIII 

m 

a 

m 
m 





6 

IX 

a 



6 

X 

p 
p 

a 



G 

XI 

P 

1 

Case  4  of  §45 

XII 

P 



P 

« 

3 

XIII 

p 

m 
m 

P 

a 



m 

m 
m 

3 

XIV 

a 



a 
a 

1 

XV 

a 

a 

P 

3 

XVI 

a 

a. 

P 

3 

XVII 

m 

1 

XVIII 

m 
m 

3 

XIX 

m 

m 

3 

XX 

p 

a 

a 

6 

XXI 

p 







2 

2 

0. 

g 

II 

c 

XXII 

p 

m 

2 

XXIII 

p 
p 

£*• 

m 





2 

XXIV 

a 

2 

XXV 

a 





m 

2 

XXVI 

p 

m 

2 

XXVII 

P 



1 

Case  2  of  §  45 

xxvirr 

a 

a 

1 

XXIX 

m 

m 

1 

21 


PROBLEM. — Determine  the  EQUILIBRANT  of  the  following  set  of  forces:    (20 
What  is  the  RESULTANT  of  the  set? 

GRAPHIC    SOLUTION 


E  A  IS  THE  REQUIRED 
EQUILIBRANT. 


PLATE  I 


1 

45°  o,o),  (15  Ibs.  90°  13,0),  (30  Ibs.  60°  7,0),  and  (25  Ibs.  315°  30,0). 

ALGEBRAIC    SOLUTION 

Let  (E   a  x,y)  be  the  force  sought,  then 
For  no  translation: 

.707  o  .500  .707 

Ax'.  E  cos  a  +  20  cos  45°  +  15  cos  90°  +  30  cos  60°  +  25  cos  315°  =  o 

+  14.14  o  +15.00  +17.68 

E  cos  a  =  —46.82 

.707  i. po  .866  —  .707 

Ay :  E  sin  a  -f  20  sin  45°  +  1 5  sin  90°  +  30  sin  60°  -f  2 5  sin  3 1 5°  =  o 
+  14.14'          +15.00  +25.98  —17.68 

E  sin  a=  —37.44 

/ano_f^^=rJi44=  8oo 

tL  cos  a       — 46.82 
a=2iS°  40' 

E  =  — s-~-  =  — -31ili  = 

sin  a   '    -.625         59'9 

For  no  rotation  : 

Taking  a  point  on  the  A-axis  for  the  point  of  application  of  E  ,  y  becomes 
zero  and  x  is  determined  as  follows : 

O  —37-44 

B :  E  cos  a  X  y  —  E  sin  ax  x  +  20  cos  45°  x  0—20  sin  45°  x  o  + 1 5  cos  90°  x  o 

o  —37.44*  ooo 

15.00  25.98  —17.68 

-i5sm9o0xi3  +  30^56o0xo-3ositt6o0x7  +  25C0s3i50xo-25-s*'tt3i 
—  195.00  o  —181.86  o  +530.40 


RESULTS 


Algebraic  Graphic 

Equilibrant  (59.9  Ibs.   218°  40'    -4.10,0)    Equilibrant  (59.9  Ibs.   218°  40'  -4.1,0) 
Resultant     (59.9  Ibs.     38°  40'    -4.10,0)    Resultant     (59.9 lbs      3 8°  40    -41,0) 


o(>  ?.c>^  g  i  -f  o 


(o,oi 
(o.o 


PROBLEM. — In  the   following   set  of  forces   in   equilibrium,    (20  Ibs.  100°  o, 
determine  the  unknowns. 

GRAPHIC    SOLUTION 


e/a 


NO  COUPLE   BEING   POSSIBLE,   THE  STRING   POLYGON    IS  SUPERFLUOUS 


Scale  i  in.  =  1 5  Ibs. 


Q-DE        P=EA 


PLATE  I  la 


2a 

60°  80° 

(15  Ibs.  150°  0,0),  (P  =  ?       or     0,0),  (10  Ibs.  340°  0,0),  and  (Q  =  ?       or     0,0), 
240°  260° 


ALGEBRAIC    SOLUTION 

—  .174  —.866  .500  .940  .174 

Ax:   20  cos  100°  +  15  cos  150°  +P  cos  6o°  +  io  cos  340°  +Q  cos  80°  =o 

—  3.48  —12.99  4  o.sooP  +9-40  +0.174(3 

.500  P-h.  1  74  (2—7.07 

.985  .500  .866  —.342  .985 

Ay:   20  sin  100°  +  15  sin  1  50°  +P  sin  60°  +  10  sin  340°  +Q  sin  80°  =o 

+7-5Q  +  0.866P  —3-42  +0.9852 


.866  P  +  .  985  0  =  23.78 

P=      32.  2  =(32.  2     60°) 
Q=  -52.6  =  (52.6   260°) 

:   No  couple  being  possible,  the  equation  of  moments  is  superfluous 


RESULTS 

Algebraic  Graphic 

(32.2lbs.     60°)  P  =  (32.3lbs.     60°) 

(52. 6  Ibs.   260°)  0^(52.5  Ibs.   260°) 


*«.<?. 


D-4- 


0 

PROBLEM — In  the  following  set  of  forces  in  equilibrium,  (20  Ibs.  270°  4,0 
determine  the  unknowns. 


GRAPHIC    SOLUTION 


Scales  i  T  !n' 

\  i  in. 


1 5  units  of  length 
15  Ibs 


P=EA    Q=DE 


PLATE    lib. 


I   2b. 


?        90°       0,0).  (10  Ibs.  90°  8,0).  (16  Ibs.  270°  14,0),  and    (Q  =  ?     90°       19,0), 
270°  270° 


ALGEBRAIC    SOLUTION 

Solution  by  .4X.  .4*.  and  B. 

00000 

/U:    20  c\9i-  270°  +P  cos  90°  +  jo  c'05  9o°  +  i6  cos  270°  +  Q  cos  90°=  o 

O  00  O  O 

—  TO  I   O  I'O  —  I    O  TO 

Av:    20  sin  270°  +P  sin  90°  +  ro  sin  90°  +  16  sm  270°  +Q  sin  90°  =o 
—  200  +P  +100  -160  +  £> 


B:      20  cos  27o°Xo  —20  sin  27o°X4  +  P  cos  9o°X  o  —  P  .nn  90°X  o  +  10  cos  90^0 

o  -I-  8  j  o  o  o 

10  o  —  T  6.0  ~  o 

—  10  szw  9o°X.8  +  1  6  6-^s  27o°Xo  —  16  si'n  27o°Xi4+Q  cos  go°Xo—Q  sin  90°  X  19=0 

—  800  o  +224.0  o  —19(2 

—  190  ==  —224.0 

Q=        ii.  8 

By/1,, 

P  =  26.0-^  =  26.0  —  1  1.  8 


Alternative  form  of  the  preceding  using  moments  alone. 
Center  of  moments  at  (o.o) 

So  -80  224 

PXo  +  20X4-  10X8+16X14+^X19=0 


Center  of  moments  at  (19,0) 


(n.8   90°) 


P  =  ^  =  14  2=  (142   90°) 

Check  by  Av: 

14.2—  20  +  10  —  i6  +  n  8=0 


RESULTS 

Algebraic  Graphic 

=  (i42ibs    90°)  P  =  (14.  2  Ibs.   90°) 

=  (ri.81bs    90C)  P  =  (i  r.8  Ibs.   90°) 


V 


A  3 


Cf 

PROBLEM. — In  the  following  set  of  forces  in  equilibrium.      (P  =  f  a  =  ?  0,0) 
mine  the  unknowns. 

GRAPHIC    SOLUTION. 


Q=? 


q     ,       {  i  in.  =15  units  of  length 
55   (i  in.  =30  Ibs. 


P=EA:  Q=DE 


PLATE   III 


:    3. 

60° 

o  Ibs.  120°  23,0),  (Q  =  ?     or       40,0).  (20  Ibs.  75°  15.  o).  and  (30  Ibs.  60°   5,0)    rlcter- 
240° 

ALGEBRAIC    SOLUTION 

Center  of  moments  at  the  point  of  application  of  the  force  which  is  other- 
wise unknown,  which,  in  this  problem,  is  the  point  (0,0), 

.866 
B:       P  cos  aXo  —  P  sin  aXo  +  10  cos  i2o°Xo  —  10  sin  i2o°X2-$+Q  cos  60° Xo 

o  o  o  — 199.18  o 

.866  .966 

—  Qsin  60°  X4o  +  20  cos  7 5° x 0—20  sin  75°X  15  +  30  cos  6o°Xo—^o  sin  60° X 5  =o 

—  34-64  Q  o  —289.80  o  —129.90 

—  34.64  Q  =618.88 

0  =  -i7-9=(i7-9   240°) 

Using  this  value  of  Q  in  the  next  two  equations, 

—  .500  17.9          —.500          .259  .500 

A»\     P  cos  a  +  10  cos  i2o°  +  Q  cos  240° +  20  cos  7 5°  +  30  cos  60°  =o 

t— 5.00  —8.85  +5.18          +15.00 

P  cos  a=  —6.23 

f  .866          17.9   ^    -.866  .966  .866 

Ay'.     P  sin  a  +  10  sin  i2o°  +  Q  sin  240° +  20  sin  75°  +  30  sin  60°  =o 

+  8.66  —15.50  +19.32  +25.98 

P  sin  a  —  =38.46 

P  sin  a      —38.46 

tan  a=-^ = — J^        =6.17 

P  cos  a      —    6.23 

a  =260°  48' 

__Pcosa_-6.23_ 
"       cos  a   ~^6o~3b-9 


RESULTvS 

Algebraic  Graphic 

P  =38.9  Ibs.     260°  48'  P  =38.9  Ibs.     261° 

Q  =  17. Q  Ibs.     240°  <2=*i7.glbs.     240° 


UNiViRsrnr  i 


>^ 


.2CfIp.fi—  0 


3Q 


PROBLEM. — In   the   following   set    of  forces    in  equilibrium,    (15    Ibs. 


and    (R     or      30,0).  Determine   P,  Q,   and/?. 
300° 


GRAPHIC    SOLUTION 


P  =  DE  Q=FA  R  =  EF 

~     .      (  i  in.  =  i  c  units  of  length 
Scales]iin.=75lbs. 


PLATE  IV. 


E    4. 

5°   0,0),    (P       or      10,0),    (20    Ibs.    90°   21,0),    Q       or      50,0),  (8  Ibs.  240°  35,0), 


330°  210° 


ALGEBRAIC    SOLUTION 

Solution  by  Ax,  Ay,  and  B. 

.707  —  866  o  .866  —.500  —  500 

Ax'-     15  COS  3I5°+P  COS   I  50° +  20  COS  90°  +  Q  COS  30° +8  COS  240°  +R  COS  I2O°=0 
10. 61  —  0.866P  o  +0.8660  —4.00  —  osoR 

—  0.866P +  0.866(2—  0.50^  =  — 6.61 


—  .707  .500  TOO  .500  —866  866 

Ay:    15  sin  3i5°+P  sin  150° +  20  sin  90°  +Q  sin  30° +  8  sin  240° +R  sin  i2o°=o 

— 10. 61     "  +o  sP  +20.00  +  0.500  —6. 93  +O.866R 

o.5oP  +  o. 50*2+0. 866^?  =  —2.46 


Center  of  Moments  at  (0,0)  o.soP 

B:     15  cos  3i5°xo-i5  sin  3i$°xo+Pcos  i$o°xo-Psin  i5o°x  10  +  20  cos  go°xo 

o  o  o  +50.P  o 

20.00  O-5O0  — 6.93 

—  2osingo°x2i+Qcos 3o°xo-Q  ^3o°x5o  +  8  Cos  24o°xo-8  sin  24 

—  420.00  o  —2S.oo(2  o  +242.55 

.866R 

+R  cos  i2o°Xo—R  sin  i2o°X3o=o 

o  —25.987? 

—  5.  ooP  —  25.00(3— 25.98^  =  177.45 


Combining  the  three  equations,  Ax,  Ay,  and  B  it  appears  that 

P  =  +6.82  =  (6.82   150°) ;  Q  =  -3.54  =  (3-54  210°) ;  R  =  -4.73  =  (4.73  300°) 


Alternative  form  of  the  preceding,  using  moments  alone: 

To  find  P,  take  center  of  moments  at  (^1,^1),  the  intersection  of  Q  and  R. 

Then  since 

y  —  -577*=  —  28-87     and     y  +  1.732^  =  51.96 
are  the  equations  of  the  lines  of  action  of  Q  and  R  respectively,  it  appears  that 

^i=35-°     and     yr=—  8.66. 
Center  of  Moments  at  (35.0,  —8.66),  transforming  co-ordinates: 

.707  —-707  —.866  .500 

15(8.66  cos  315°  --  35  sin  315°)  +P(8.66  cos  150°  --  25  sin  150°) 

+  6.12  —24.75  —7-5°  +12.50 

o  i.oo  —  .500  —  .866 

+  20(8.66  cos  90°  --  145^90°)  +8(8.66  cos  240°—  o  sin  240°)  +QXo 

o  14.00  —433  o 


5.  oP  =  279.  45  —280.00  +  34.64=34.09.  Whence  P=  +6.82  =(6.82  150°) 
To  find  Q,  take  center  of  moments  at  intersection  of  P  and  R  and  proceed 
as  for  P. 

Similarly,  to  find  R,  take  center  of  moments  at  the  intersection  P  and  Q. 
In  practical  cases  the  algebraic  work  usually  proves  much  simpler  than 
for  the  general  case  used  in  this  Plate.     See  Plate  V. 

RESULTS 

Algebraic  Graphic 

P  =(6.  82  Ibs.    150°)  P  =  (6.81bs.    150°) 

Q  =  (3.54lbs.   210°)  Q  =  (3.61bs.   210°) 

R  =  (4.73  Ibs.  300°)  /2  =  (4.7lbs.  300°) 


V^  ^fjuo 

t.|.-  -*A  ;(°ei£  i-^*44££2^^:(co?  r 


atrft  ni 


k\\ 


•'  A3-  p  : 


CASE  4  (secc 


PROBLEM. — A  body  in  the  shape  of  an  isosceles  triangle,  base  20  feet,  altitude 
are  known  in  line  of  action  only.      Determine  P,  Q,  and  R. 


GRAPHIC    SOLUTION 


P-DE  Q=EA  R=CD 


=  ^oo  Ibs. 


PLATE  V. 


I  example). 

feet,  is  in   equilibrium  under  the  action  of  the  forces  shown;  of  these  P,  Q,  and  R 


ALGEBRAIC    SOLUTION 


Center  of  moments  at  intersection  of  P  and  Q 
7^X15  +600X10—650X20+^X0+^X0=0 

6000  13000 

600X10—6(50X20      7000 

—  =  —  -  =  +466. 67  =(466. 67    1 80°) 
15  15 


Center  of  moments  at  intersection  of  Q  and  R 

PX  15  cos  26°  341+6ooXio-65oX2o+(2Xo+7?Xo=o 

6coo  13000 

600X10—650X20       7000 

-"-  --— --5».o-(5».o  333-*') 


Center  of  moments  at  intersection  of  P  and  R 
(3X30— 


600X20—6COXIO  55OO 

~  -  =  +183-33  =(183.  33   90°) 

30  30 


RESULTS 

Algebraic  Graphic 

P  =  (522.0  Ibs.   333°26')  P  =  (522  Ibs.   33o°26') 

Q  =  (183.3  Ibs.     9°°  o')  Q  =  (183  Ibs.     90°  o') 

R  =  (466. 7  Ibs.   180°  o')  R  =  (467  Ibs.    180°  o') 


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Drechsel's  Chemical  Reactions.     (Merrill.) i2mo,  i  25 

Duhem's  Thermodynamics  and  Chemistry.     (Burgess.)     (Shortly.) 

Eissler's  Modern  High  Explosives 8vo,  4  oo 

3 


Effront's  Enzymes  and  their  Applications.     (Prescott.) 8vo,  3  oo 

Erdmann's  Introduction  to  Chemical  Preparations.     (Dunlap.) i2mo,  i  25 

Fletcher's  Practical  Instructions  in  Quantitative  Assaying  with  the  Blowpipe. 

lamo,  morocco,  i   50 

Fowler's  Sewage  Works  Analyses i2mo,  2  oo 

Fresenius's  Manual  of  Qualitative  Chemical  Analysis.     (Wells.) 8vo,  5  oo 

Manual  of  Qualitative  Chemical  Analysis.     Parti.    Descriptive.     (Wells.) 

8vo,  3  oo 

System   of   Instruction   in    Quantitative    Chemical   Analysis.      (Cohn.) 
2vols.    (Shortly.) 

Fuertes's  Water  and  Public  Health i2mo,  i  50 

Furman's  Manual  of  Practical  Assaying 8vo,  3  oo 

Gill's  Gas  and  Fuel  Analysis  for  Engineers i2mo,  i  25 

Grotenfelt's  Principles  of  Modern  Dairy  Practice.     (Wo  11.) i2mo.  2  oo 

Hammarsten's  Text-book  of  Physiological  Chemistry.     (Mandel.) 8vo,  4  oo 

Helm's  Principles  of  Mathematical  Chemistry.     (Morgan.) i2mo.  i  50 

Hinds's  Inorganic  Chemistry 8vo,  3  oo 

•  Laboratory  Manual  for  Students i2mo,  75 

Ho  lie  man's  Text-book  of  Inorganic  Chemistry.     (Cooper.) 8vo,  2  50 

Text-book  of  Organic  Chemistry.     (Walker  and  Mott.) 8vo,  2  50 

Hopkins's  Oil-chemists'  Handbook 8vo,  3  oo 

Jackson's  Directions  for  Laboratory  Work  in  Physiological  Chemistry.  .8vo,  i  oo 

Keep's  Cast  Iron 8vo,  2  50 

Ladd's  Manual  of  Quantitative  Chemical  Analysis i2mo.  i  oo 

Landauer's  Spectrum  Analysis.    (Tingle.) 8vo,  3  oo 

Lassar-Cohn's  Practical  Urinary  Analysis.     (Lorenz.) i  amo.  i  oo 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State 

Control.     (In  preparation.) 

Lob's  Electrolysis  and  Electrosynthesis  of  Organic  Compounds.  (Lorenz.)  i2mo,  i  oo 

Mandel's  Handbook  for  Bio-chemical  Laboratory i2mo,  i  50 

Mason's  Water-supply.     (Considered  Principally  from  a  Sanitary  Standpoint.) 

3d  Edition,  Rewritten 8vo,  4  oo 

Examination  of  Water.     (Chemical  and  Bacteriological.) 1 2ino,  i  25 

Meyer's  Determination  of  Radicles  in  Carbon  Compounds.     (Tingle.).  .i2mo,  i  oo 

Miller's  Manual  of  Assaying i2mo,  i  oo 

Milter's  Elementary  Text-book  of  Chemistry i2mo,  i  50 

Morgan's  Outline  of  Theory  of  Solution  and  its  Results 12 mo,  i  oo 

Elements  of  Physical  Chemistry I2mo.  2  oo 

Nichols's  Water-supply.     (Considered  mainly  from  a  Chemical  and  Sanitary 

Standpoint,  1883.) 8vo,  2  50 

O'Brine's  Laboratory  Guide  in  Chemical  Analysis 8vo,  2  oo 

O'Driscoll's  Notes  on  the  Treatment  of  Gold  Ores 8vo,  2  oo 

Ost  and  Kolbeck's  Text-book  of  Chemical  Technology.     (Lorenz — Bozart.) 
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*  Penfield's  Notes  on  Determinative  Mineralogy  and  Record  of  Mineral  Tests. 

8vo,  paper,  50 

Pictet's  The   Alkaloids  and  their  Chemical  Constitution.      (Biddle.)      (In 
preparation.) 

Pinner's  Introduction  to  Organic  Chemistry.     (Austen.) i2mo,  i  50 

Poole's  Calorific  Power  of  Fuels 8vo»  3  oo 

*  Reisig's  Guide  to  Piece-dyeing 8vo,  25  oo 

Richards  and  Woodman's  Air  .Water ,  and  Food  from  a  Sanitar  ^Standpoint .  8vo,  2  oo 

Richards' s  Cost  of  Living  as  Modified  by  Sanitary  Science i2mo,  i  oo 

Cost  of  Food  a  Study  in  Dietaries i2mo,  i  oo 

•  Richards  and  Williams's  The  Dietary  Computer 8vo,  i  50 

Ricketts  and  Russell's  Skeleton  Notes  upon  Inorganic  Chemistry.     (Part  I. — 

Non-metallic  Elements.) 8vo,  morocco,  75 

Ricketts  and  Miller's  Notes  on  Assaying 8vo,  3  oo 


d  eal's  Sewage  and  the  Bacterial  Purification  of  Sewage 8vo,  3  50 

Ruddiman's  Incompatibilities  in  Prescriptions 8vo,  a  oo 

Schimpf's  Text-book  of  Volumetric  Analysis 121110.  2  50 

Spencer's  Handbook  for  Chemists  of  Beet-sugar  Houses i6mo,  morocco,  3  oo 

Handbook  for  Sugar  Manufacturers  and  their  Chemists. .  i6mo,  morocco,  2  oo 

Stockbridge's  Rocks  and  Soils 8vo,  2  50 

*  Tillman's  Elementary  Lessons  in  Heat 8vo,  i  50 

*  Descriptive  General  Chemistry 8vo  3  oo 

Treadwell's  Qualitative  Analysis.     (Hall.) 8vo,  3  oo 

Turneaure  and  Russell's  Public  Water-supplies 8vo,  5  oo 

Van  Deventer's  Physical  Chemistry  for  Beginners.     (Boltwood.) 121110,  i  50 

*  Walke's  Lectures  on  Explosives 8vo,  4  oo 

Wells's  Laboratory  Guide  in  Qualitative  Chemical  Analysis 8vo,  i  so 

Short  Course  in  Inorganic  Qualitative  Chemical  Analysis  for  Engineering 

Students lamo,  i  50 

Whipple's  Microscopy  of  Drinking-water 8vo,  3  50 

Wiechmann's  Sugar  Analysis Small  8vo,  2  50 

Wilson's  Cyanide  Processes i2mo,  i  50 

Chlorination  Process i2mo,  i  50 

Wulling's  Elementary  Course  in  Inorganic  Pharmaceutical  and  Medical  Chem- 
istry  i2mo,  2  oo 

CIVIL  ENGINEERING. 

BRIDGES  AND    ROOFS.       HYDRAULICS.      MATERIALS    OF  ENGINEERING. 
RAILWAY  ENGINEERING. 

Baker's  Engineers'  Surveying  Instruments 12 mo,  3  oo 

Bixby's  Graphical  Computing  Table Paper,  10.^X241  inches  25 

**  Burr's  Ancient  and  Modern  Engineering  and  the  Isthmian  Canal.     (Postage 

27  cents  additional.) 8vo,  m  3  50 

Comstock's  Field  Astronomy  for  Engineers 8vo,  z  50 

Davis's  Elevation  and  Stadia  Tables 8vo,  I  oo 

Elliott's  Engineering  for  Land  Drainage i2mo,  i  50 

Practical  Farm  Drainage i2mo,  I  oo 

Folwell's  Sewerage.     (Designing  and  Maintenance.) 8vo,  3  oo 

Preitag's  Architectural  Engineering.     2d  Edition,  Rewritten 8vo,  3  50 

French  and  Ives's  Stereotomy 8vo,  a  50 

Goodhue's  Municipal  Improvements i2mo,  i  75 

Goodrich's  Economic  Disposal  of  Towns'  Refuse 8vo,  3  50 

Gore's  Elements  of  Geodesy 8vo,  a  50 

Hayford's  Text-book  of  Geodetic  Astronomy 8vo,  3  oo 

Howe's  Retaining  Walls  for  Earth i2mo,  i  as 

Johnson's  Theory  and  Practice  of  Surveying Small  8vo.  4  oo 

Statics  by  Algebraic  and  Graphic  Methods 8vo,  a  oo 

Kiersted's  Sewage  Disposal i2mo,  i  25 

Laplace's  Philosophical  Essay  on  Probabilities.     (Truscott  and  Emory.)  12010,  a  oo 

Mahan's  Treatise  on  Civil  Engineering.    (1873.)    (Wood.) 8vo0  5  oo 

*  Descriptive  Geometry 8vo,  i  50 

Merriman's  Elements  of  Precise  Surveying  and  Geodesy 8vo,  a  50 

Elements  of  Sanitary  Engineering 8vo,  a  oo 

Merriman  and  Brooks's  Handbook  for  Surveyors i6mo,  morocco,  2  oo 

Nugent's  Plane  Surveying 8vo,  3  50 

Ogden's  Sewer  Design i2mo,  2  oo 

Patton's  Treatise  on  Civil  Engineering 8vo,  half  leather,  7  SO 

Reed's  Topographical  Drawing  and  Sketching 4to,  5  oo 

Rideal'sJSewage  and  the  Bacterial  Purification  of  Sewage 8vo,  3  50 

Siebert  and  Biggin's  Modern  Stone-cutting  and  Masonry 8vo,  i  50 

Smith's  Manual  of  Topographical  Drawing.     (McMillan.) 7 .8vo,  a  50 

5 


Sondericker's  Graphic   Statics,   witn   ^-pplications   to   Trusses,  Beams,  and 
Arches.     (Shortly.) 

*  Trantwine's  Civil  Engineer's  Pocket-book i6mo,  morocco,  5  oo 

Wait's  Engineering  and  Architectural  Jurisprudence 8vo,  6  oo 

Sheep,  6  50 

Law  of  Operations  Preliminary  to  Construction  in  Engineering  and  Archi- 
tecture.   8vo,  5~~oo 

10  i  Sheep,  5  5<> 

Law  of  Contracts 8vo,  3  oo 

Warren's  Stereotomy — Problems  in  Stone-cutting 8vo,  2  50 

Webb's  Problems  in  the  U«e  and  Adjustment  of  Engineering  Instruments. 

i6mo,  morocco,  i  25 

*  Wheeler's  Elementary  Course  of  Civil  Engineering 8vo,  4  oo 

Wilson's  Topographic  Surveying 8vo,  3^50 


BRIDGES  AND  ROOFS. 

Boiler's  Practical  Treatise  on  the  Construction  of  Iron  Highway  Bridges.  . 8vo,  2  oo 

*         Thames  River  Bridge 4to,  paper,  5  oo 

Burr's  Course  on  the  Stresses  in  Bridges  and  Roof  Trusses,  Arched  Ribs,  and 

Suspension  Bridges 8vo,  3  50 

Du  Bois's  Mechanics  of  Engineering.     Vol.  II Small  4to,  10  oo 

Poster's  Treatise  on  Wooden  Trestle  Bridges 4to,  5  oo 

Fowler's  Coffer-dam  Process  for  Piers 8vo,  2  50 

Greene's  Roof  Trusses 8vo,  i  25 

Bridge  Trusses 8vo,  2  50 

Arches  in  Wood,  Iron,  and  Stone 8vo,  2  50 

Howe's  Treatise  on  Arches ^ 8vo  4  oo 

Design  of  Simple  Roof-trusses  in  Wood  and  Steel 8vo,  2  oo 

Johnson,  Bryan,  and  Turneaure's  Theory  and  Practice  in  the  Designing  of 

Modern  Framed  Structures Small  4to,  10  oo 

Merriman  and  Jacoby's  Text-book  on  Roofs  and  Bridges: 

Part  I. — Stresses  in  Simple  Trusses 8vo,  2  50 

Part  H. — Graphic  Statics 8vo,  2  50 

Part  III. — Bridge  Design.     4th  Edition,  Rewritten 8vo,  2  50 

Part  IV. — Higher  Structures 8vo,  2  50 

Morison's  Memphis  Bridge 4to,  10  oo 

Waddell's  De  Pontibus,  a  Pocket-book  for  Bridge  Engineers. . .  i6mo,  morocco,  3  oo 

Specifications  for  Steel  Bridges i2mo,  i  25 

Wood's  Treatise  on  the  Theory  of  the  Construction  of  Bridges  and  Roofs. 8vo,  2  oo 
Wright's  Designing  of  Draw-spans: 

Part  I.  — Plate-girder  Draws 8vo,  2  50 

Part  II. — Riveted-truss  and  Pin-connected  Long-span  Draws 8vo,  2  50 

Two  parts  in  one  volume 8vo,  3  50 


HYDRAULICS. 

Bazin's  Experiments  upon  the  Contraction  of  the  Liquid  Vein  Issuing  from  an 

Orifice.     (Trautwine.) 8vo,  2  oo 

Bovey's  Treatise  on  Hydraulics .' 8vo,  5  oo 

Church's  Mechanics  of  Engineering 8vo,  6  oo 

Diagrams  of  Mean  Velocity  of  Water  in  Open  Channels paper,  i  50 

Coffin's  Graphical  Solution  of  Hydraulic  Problems i6mo,  morocco,  2  50 

Flather's  Dynamometers,  and  the  Measurement  of  Power 12010,  3  oo 

Folwell's  Water-supply  Engineering. 8vo,  4  oo 

Frizell's  Water-power 8vo,  5  oo 


Fuertes's  Water  and  Public  Health lamo,    i  50 

Water-filtration  Works i2mo,    2  50 

Ganguillet  and  Kutter's  General  Formula  for  the  Uniform  Flow  of  Water  in 

Rivers  and  Other  Channels.     (Hering  and  Trautwine.)-^ 8vo,    4  oo 

Hazen's  Filtration  of  Public  Water-supply 8vo,    3  oo 

Hazlehurst's  Towers  and  Tanks  for  Water- works 8vo,    2  50 

Herschel's  115  Experiments  on  the  Carrying  Capacity  of  Large,  Riveted,  Metal 

Conduits 8vo,    2  oo 

Mason's    Water-supply.     (Considered    Principally   from   a    Sanitary   Stand- 
point.)    3d  Edition,  Rewritten 8vo,  4  oo 

Merriman's  Treatise  on  Hydraulics,     gth  Edition,  Rewritten 8vo,    5  oo 

*  Michie's  Elements  of  Analytical  Mechanics 8vo,    4  oo 

Schuyler's  Reservoirs  for  Irrigation,  Water-power,  and  Domestic   Water- 
supply Large  8vo,    5  oo 

**  Thomas  and  Watt's  Improvement  of  Riyers.     (Post.,  44  c.  additional),  4to,    6  oo 

Turneaure  and  Russell's  Public  Water-supplies 8vo.    5  oo 

Wegmann's  Design  and  Construction  of  Dams 4to,    5  oo 

Water-suoolv  of  the  City  of  New  York  from  1658  to  1895 4to,  10  oo 

Weisbach's  Hydraulics  and  Hydraulic  Motors.     (Du  Bois.) 8vo,    5  oo 

Wilson's  Manual  of  Irrigation  Engineering Small  8vo,    4  oo 

Wolff's  Windmill  as  a  Prime  Mover 8vo,'  3  oo 

Wood's  Turbines 8vo,    a  50 

Elements  of  Analytical  Mechanics 8vo,    3  oo 


MATERIALS  OF  ENGINEERING. 

Baker's  Treatise  on  Masonry  Construction 8vo,    5  oo 

Roads  and  Pavements 8vo,    5  oo 

Black's  United  States  Public  Works Oblong  4to,    5  oo 

Bovey's  Strength  of  Materials  and  Theory  of  Structures 8vo,    7  50 

Burr's  Elasticity  and  Resistance  of  the  Materials  of  Engineering.     6th  Edi- 
tion, Rewritten 8vo,    7  50 

Byrne's  Highway  Construction 8vo,    5  oo 

Inspection  of  the  Materials  and  Workmanship  Employed  in  Construction. 

i6mo,    3  oo 

Church's  Mechanics  of  Engineering 8vo,    6  oo 

Du  Bois's  Mechanics  of  Engineering.     Vol.  I Small  4to,    7  50 

Johnson's  Materials  of  Construction Large  8vo,    6  oo 

Keep's  Cast  Iron 8vo,    2  50 

Lanza's  Applied  Mechanics 8vo,   7  50 

Martens's  Handbook  on  Testing  Materials.     (Henning.)     2  vols 8vo,    750 

Merrill's  Stones  for  Building  and  Decoration 8vo,    5  oo 

Merriman's  Text-book  on  the  Mechanics  of  Materials 8vo,    4  oo 

Strength  of  Materials i  amo,    i  oo 

Metcalf's  Steel.     A  Manual  for  Steel-users i2mo,    2  oo 

Patton's  Practical  Treatise  on  Foundations 8vo,    5  oo 

Rockwell's  Roads  and  Pavements  in  France i2mo,    i  25 

Smith's  Wire :  Its  Use  and  Manufacture Small  4to,   3  oo 

Materials  of  Machines .'. .  i2mo,    i  oo 

Snow's  Principal  Species  of  Wood 8vo,    3  50 

Spalding's  Hydraulic  Cement i2mo,    2  oo 

Text-book  on  Roads  and  Pavements i2mo,    2  oo 

Thurston's  Materials  of  Engineering.     3  Parts 8vo,    8  oo 

Part  I. — Non-metallic  Materials  of  Engineering  and  Metallurgy 8vo,    2  oo 

-    Part  II. — Iron  and  Steel 8vo,   3  50 

Part  III. — A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents 8vo,    2^50 

7 


Thurston's  Text-book  of  the  Materials  of  Construction 8vo,  5  oo 

Tillson's  Street  Pavements  and  Paving  Materials 8vo,  4  oo 

Waddell's  De  Pontibus.     (A  Pocket-book  for  "Bridge  Engineers.) .  .  i6mo,  mor.,  3  oo 

Specifications  for  Steel  Bridges i2mo,  i  23 

Wood's  Treatise  on  the  Resistance  of  Materials,  and  an  Appendix  on  the  Pres- 
ervation of  Timber 8vo,  2  oo 

Elements  of  Analytical  Mechanics 8vo,  3  oo 


RAILWAY  ENGINEERING. 

Andrews's  Handbook  for  Street  Railway  Engineers.     3X5  inches,  morocco,  i  25 

Berg's  Buildings  and  Structures  of  American  Railroads 4to,  5  oo 

Brooks's  Handbook  of  Street  Railroad  Location i6mo.  morocco,  i  50 

Butts's  Civil  Engineer's  Field-book i6mo,  morocco,  2  50 

Crandall's  Transition  Curve i6mo,  morocco,  i  50 

Railway  and  Other  Earthwork  Tables 8vo,  i  50 

Da wson's  "Engineering"  and  Electric  Traction  Pocket-book.    i6mo,  morocco,  4  oo 

Dredge's  History  of  the  Pennsylvania  Railroad:   (1879) Paper,  5  oo 

*  Drinker's  Tunneling,  Explosive  Compounds,  and  Rock  Drills,  4to,  half  mor.,    25  oo 

Fisher's  Table  of  Cubic  Yards Cardboard,  25 

Godwin's  Railroad  Engineers'  Field-book  and  Explorers'  Guide i6mo,  mor.,  2  50 

Howard's  Transition  Curve  Field-book i6mo  morocco  i  50 

Hudson's  Tables  for  Calculating  the  Cubic  Contents  of  Excavations  and  Em- 
bankments    8vo,  i  oo 

Mo  lit  or  and  Beard's  Manual  for  Resident  Engineers i6mo,  i  oo 

Nagle's  Field  Manual  for  Railroad  Engineers 1 6mo,  morocco.  3  oo 

Philb rick's  Field  Manual  for  Engineers i6mo,  morocco,  3  oo 

Pratt  and  Alden's  Street-railway  Road-bed 8vo,  2  oo 

Searles's  Field  Engineering i6mo,  morocco,  3  oo 

Railroad  Spiral. i6mo,  morocco  i  50 

Taylor's  Prismoidal  Formulae  and  Earthwork 8vo,  i  50 

*  Trautwine's  Method  of  Calculating  the  Cubic  Contents  of  Excavations  and 

Embankments  by  the  Aid  of  Diagrams 8vo,  2  oo 

he  Field  Practice  of  [Laying   Out    Circular    Curves   for    Railroads. 

i2mo,  morocco,  2  50 

*  Cross-section  Sheet Paper,  25 

Webb's  Railroad  Construction.     2d  Edition,  Rewritten 16010.  morocco,  5  oo 

Wellington's  Economic  Theory  of  the  Location  of  Railways Small  8vo,  5  oo 


DRAWING. 

Barr's  Kinematics  of  Machinery 8vo,  2  50 

*  Bartlett's  Mechanical  Drawing 8vo,  3  oo 

Coolidge's  Manual  of  Drawing 8vo,  paper,  i  oo 

Durley's  Kinematics  of  Machines 8vo,  4  oo 

Hill's  Text-book  on  Shades  and  Shadows,  and  Perspective 8vo,  2  oo 

Jones's  Machine  Design: 

Part  I. — Kinematics  of  Machinery 8vo,  i  50 

Part  H. — Form,  Strength,  and  Proportions  of  Parts 8vo,  3  oo 

MacCord's  Elements  of  Descriptive  Geometry 8vo,  3  oo 

Kinematics;  or,  Practical  Mechanism 8vo,  5  oo 

Mechanical  Drawing 4to,  4  oo 

Velocity  Diagrams 8vo,  i  50 

*  Mahan's  Descriptive  Geometry  and  Stone-cutting 8vo,  i  50 

Industrial  Drawing.    (Thompson.) 8vo,  3  50 

Reed's  Topographical  Drawing  and  Sketching 4to,  5  oo 

8 


Reid's  Course  in  Mechanical  Drawing 8vo,  2  oo 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design..  8  vo,  3  oo 

Robinson's  Principles  of  Mechanism 8vo,  3  oo 

Smith's  Manual  of  Topographical  Drawing.     (McMillan.) 8vo,  2  50 

Warren's  Elements  of  Plane  and  Solid  Free-hand  Geometrical  Drawing. .  12 mo, 


Drafting  Instruments  and  Operations i2mo, 

Manual  of  Elementary  Projection  Drawing 1 2mo, 

Manual  of  Elementary  Broblems  in  the  Linear  Perspective  of  Form  and 


Shadow i2mo,  oo 

Plane  Problems  in  Elementary  Geometry i2mo,  25 

Primary  Geometry i2mo,  75 

Elements  of  Descriptive  Geometry,  Shadows,  andJPerspective 8vo,  3  50 

General  Problems  of  Shades  and  Shadows 8vo,  3  oo 

Elements  of  Machine  Construction  and  Drawing 8vo,  7  So 

Problems.  Theorems,  and  Examples  in  Descriptive  Geometrv 8vo,  2  50 

Weisbach's  Kinematics  and  the  Power  of  Transmission.       (Hermann  an'' 

Klein.) 8vo,  5  oo 

Whelpley's  Practical  Instruction  in  the  Art  of  Letter  Engraving i2mo,  2  oo 

Wilson's  Topographic  Surveying 8vo,  3  50 

Free-hand  Perspective 8vo,  2  50 

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Woolf's  Elementary  Course  in  Descriptive  Geometry Large  8vo,  3  oo 


"ELECTRICITY  AND   PHYSICS. 

Anthony  and  Brackett's  Text-book  of  Physics.  (Magie.) ....  .Small  8vo,  3  oo 

Anthony's  Lecture-notes  on  the  Theory  of  Electrical  Measurements i2mo,  i  oo 

Benjamin's]History  of  Electricity 8vo,  3  oo 

Voltaic  CelL 8vo,  3  oo 

Classen's  Quantitative  Chemical  Analysis  by  Electrolysis.  (Boltwood.).  .8vo,  3  oo 

Crehore  and  Squier's  Polarizing  Photo-chronograph 8vo,  3  oo 

Dawson's  "Engineering"  and  Electric  Traction  Pocket-book. .  lomo,  morocco,  4  oo 

Flather's  Dynamometers,  and  the  Measurement  of  Power 121110,  3  oo 

Gilbert's  De  Magnete.  (Mottelay.) 8vo,  2  50 

Holman's  Precision  of  Measurements 8vo,  2  oo 

Telescopic  Mirror-scale  Method,  Adjustments,  and  Tests Large  8vo  75 

Lanaauer's  Spectrum  Analysis.  (Tingle-.) 8vo,  3  oo 

Le  Chatelier's  High- temperature  Measurements.  (Boudouard — Burgess.  )i2mo,  3  oo 

Lob's  Electrolysis  and  Electrosynthesis  of  Organic  Compounds.  (Lorenz.)  i2mo,  i  oo 

*  Lyons's Treatise  on  Electromagnetic  Phenomena.    Vols.  I.  and  II.  8vo,  each,*  61  oo 

*  Michie.     Elements  of  Wave  Motion  Relating  to|Soundjand  Light 8vp,^£  °-° 

Niaudet's  Elementary  Treatise  on  Electric  Batteries.     (FishoacK. ) i2mo,  2  TO 

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*  Rosenberg's  Electrical  Engineering.    (Haldane  Gee — Kinzbrunner.) 8vo,  i   50 

Ryan,  Norris,  and  Hozie's  Electrical  Machinery.     (In  preparation.' 

Thurston's  Stationary  Steam-engines 8vo,  2  50 

*  Tillman's  Elementary  Lessons  in  Heat 8vo,  i  50 

Tory  and  Pitcher's  Manual  of  Laboratory  Physics Small  8vo,  2  oo 

Hike's  Modern  Electrolytic  Copper  Refining 8vo,  3  oo 


LAW. 

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*  Treatise  on  the  Military  Law  of  United  States 8vo,  7  oo 

*  Sheep,  7  50 
Manual  for  Courts-martial i6mo,  morocco,  i  50 


Wait's  Engineering  and  Architectural  Jurisprudence 8vo,  6  oo 

Sheep,  6  50 

Law  of  Operations  Preliminary  to  Construction  in  Engineering'and  Archi- 
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Sheep,  5  50 

Law  of  Contracts 8vo,  3  oo 

Winthrop's  Abridgment  of  Military  Law I2mo,  a  50 


MANUFACTURES. 

Bernadou's  Smokeless  Powder — Nitro-cellulose  and  Theory  of  the  Cellulose 

Molecule i amo,  a  50 

Holland's  Iron  Founder i2mo,  2  50 

"  The  Iron  Founder,"  Supplement. i2mo,  2  50 

Encyclopedia  of  Founding  and  Dictionary  of  ^Foundry  Terms  Used  in  the 

Practice  of  Moulding iamo,  3  oo 

Eissler's  Modern  High  Explosives 8vo,  4  oo 

Eff rent's  Enzymes  and  their  Applications.     (Prescott.) 8vo,  3  oo 

Fitzgerald's  Boston  Machinist i8mo,  i  oo 

Ford's  Boiler  Making  for  Boiler  Makers i8mo,  i  oo 

Hopkins's  Oil-chemists'  Handbook 8vo,  3  oo 

Keep's  Cast  Iron 8vo,  a  50 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special_Ref erence  to  State 

Control.     (In  preparation.) 

Metcalf's  SteeL    A  Manual  for  Steel-users lamo,  a  oo 

Metcalfe's  Cost  of  Manufactures — And  the  Administration    of  Workshops, 

Public  and  Private 8vo,  5  oo 

Meyer's  Modern  Locomotive  Construction 4to,  10  oo 

*  Reisig's  Guide  to  Piece-dyeing 8vo,  25  oo 

Smith's  Press-working  of  Metals 8vo,  3  oo 

Wire:  Its  Use  and  Manufacture Small  4to,  3  oo 

Spalding's  Hydraulic  Cement i2mo,  2  oo 

Spencer's  Handbook  for  Chemists  of  Beet-sugar  Houses i6mo,  morocco,  3  oo 

Handbook  tor  sugar  Manufacturers  and  their  Chemists.. .  i6mo,  morocco,  2  oo 
Thurston's  Manual  of  Steam-boilers,  their  Designs,  Construction  and  Opera- 
tion  8vo,  5  oo 

*  Walke's  Lectures  on  Explosives 8vo,  4  oo 

West's  American  Foundry  Practice xarno,  a  50 

Moulder's  Text-book iamo,  a  50 

Wiechmann's  Sugar  Analysis Small  8vo,  a  50 

Wolff's  Windmill  as  a  Prime  Mover 8vo,  3  oo 

Woodbury's  Fire  Protection  of  Mills 8vo,  a  50 


MATHEMATICS. 

Baker's  Elliptic  Functions 8vo,    i  50 

*  Bass's  Elements  of  Differential  Calculus lamo,    4  oo 

Briggs's  Elements  of  Plane  Analytic  Geometry tamo, 

Chapman's  Elementary  Course  in  Theory  of  Equations i2mo, 

Compton's  Manual  of  Logarithmic  Computations 12 mo, 

Davis's  Introduction  to  the  Logic  of  Algebra 8vo, 

*  Dickson's  College  Algebra Large  i2mo, 


*       Introduction  to  the  Theory  of  Algebraic  Equations   LargeTi2mo, 

Halsted's  Elements  of  Geometry 8vo, 

Elementary  Synthetic  Geometry 8vo. 

10 


oo 
50 
50 
50 
50 
25 
75 
50 


*  Johnson's  Three-place  Logarithmic  Tables:    Vest-pocket  size paper,  '     15 

100  copies  for  5  oo 

*  Mounted  on  heavy  cardboard,  8  X 10  inches,  25 

10  copies  for  2  oo 

Elementary  Treatise  on  the  Integral  Calculus Small  8vo,  i  50 

Curve  Tracing  in  Cartesian  Co-ordinates I2mo,  I  oo 

Treatise  on  Ordinary  and  Partial  Differential  Equations Small  8vo,  3  5<> 

Theory  of  Errors  and  the  Method  of  Least  Squares i2mo,  I  50 

*  Theoretical  Mechanics i2mo,  3  oo 

Laplace's  Philosophical  Essay  on  Probabilities.     (Truscott  and  Emory.)  i2mo,  2  oo 

*  Ludlow  and  Bass.     Elements  of  Trigonometry  and  Logarithmic  and  Other 

Tables 8vo,  3  oo 

Trigonometry  and  Tables  published  separately Each,    2  oo 

Maurer's  Technical  Mechanics.     (In  preparation.) 

Merriman  and  Woodward's  Higher  Mathematics 8vo,  5  oo 

Merriman's  Method  of  Least  Squares 8vo,  2  oo 

Rice  and  Johnson's  Elementary  Treatise  on  the  Differential  Calculus .  Sm.,  8vo,  3  oo 

Differential  and  Integral  Calculus.     2  vols.  in  one Ginall  8vo,  2  50 

Wood's  Elements  of  Co-ordinate  Geometry 8vo,  2  oo 

Trigonometry:  Analytical,  Plane,  and  Spherical iamo,  i  oo 

MECHANICAL   ENGINEERING. 
MATERIALS  OF  ENGINEERING,  STEAM-ENGINES  AND  BOILERS. 

Baldwin's  Steam  Heating  for  Buildings i2mo,  2  50 

Barr's  Kinematics  of  Machinery 8vo,  2  50 

*  Bartlett's  Mechanical  Drawing 8vo,  3  oo 

Benjamin's  Wrinkles  and  Recipes I2mo,  2  oo 

Carpenter's  Experimental  Engineering 8vo,  6  oo 

Heating  and  Ventilating  Buildings 8vo,  4  oo 

Clerk's  Gas  and  Oil  Engine Small  8vo,  4  oo 

Coolidge's  Manual  of  Drawing 8vo,    paper,  i  oo 

Cromwell's  Treatise  on  Toothed  Gearing i2mo,  i  50 

Treatise  on  Belts  and  Pulleys i2mo,  i  50 

Durley's  Kinematics  of  Machines 8vo,  4  oo 

Flather's  Dynamometers  and  the  Measurement  of  Power 12 mo,  3  oo 

Rope  Driving i2mo,  2  oo 

Gill's  Gas  and  Fuel  Analysis  for  Engineers 12010,  i  25 

Hall's  Car  Lubrication i2mo,  i  oo 

Button's  The  Gas  Engine.     (In  preparation.) 
Jones's  Machine  Design: 

Part  I.— Kinematics  of  Machinery 8vo,  i  50 

Part  II. — Form,  Strength,  and  Proportions  of  Parts 8vo,  3  oo 

Kent's  Mechanical  Engineer's  Pocket-book i6mo,    morocco,  5  oo 

Kerr's  Power  and  Power  Transmission 8vo,  2  oo 

Mac  Cord's  Kinematics;  or,  Practical  Mechanism 8vo,  5  oo 

Mechanical  Drawing 4to,  4  oo 

Velocity  Diagrams 8vo,  i  50 

Mahan's  Industrial  Drawing.    (Thompson.) 8vo,  3  So 

Poole's  Calorific  Power  of  Fuels 8vo,  3  oo 

Reid's  Course  in  Mechanical  Drawing 8vo.  2  oo 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design . .  8vo ,  3  oo 

Richards's  Compressed  Air I2mo,  i  50 

Robinson's  Principles  of  Mechanism 8vo,  3  oo 

Smith's  Press-working  of  Metals -^8vo  3  oo 

Thurston's  Treatise  on   Friction  and    Lost  Work  in   Machinery  and   Mil 

Work 8vo,  3  oo 

Animal  as  a  Machine  and  Prime  Motor,  and  the  Laws  of  Energetics .  i2mo,  i  oo 

11 


Warren's  Elements  of  Machine  Construction  and  Drawing 8  vo,  7  50 

Weisbach's  Kinematics  and  the  Power  of  Transmission.      Herrmann- 
Klein.) 8vo,  5  ~oo 

Machinery  of  Transmission  and  Governors.     (Herrmann — Klein.).  .8vo,  500 

Hydrauhcs  and  Hydraulic  Motors.     (Du  Bois.) 8vo,  5  oo 

Wolff's  Windmill  as  a  Prime  Mover 8vo,  3  oo 

Wood's  Turbines 8vo,  2  50 

MATERIALS  OF  ENGINEERING. 

Bovey's  Strength  of  Materials  and  Theory  of  Structures 8vo,  7  50 

Burr's  Elasticity  and  Resistance  of  the  Materials  of  Engineering.     6th  Edition, 

Reset 8vo,  7  50 

Church's  Mechanics  of  Engineering 8vo,  6  oo 

Johnson's  Materials  of  Construction Large  8vo,  6  oo 

Keep's  Cast  Iron 8vo  2  50 

Lanza's  Applied  Mechanics 8vo,  7  50 

Martens's  Handbook  on  Testing  Materials.     (Henning.) 8vo,  7  50 

Merriman's  Text-book  on  the  Mechanics  of  Materials 8vo,  4  oo 

Strength  of  Mater (als 1 2  mo ,  i  oo 

Metcalf's  SteeL     A  Manual  for  Steel-users i2mo  2  oo 

Smith's  Wire:  Its  Use  and  Manufacture Small  4to,  3  oo 

Materials  of  Machines i2mo,  i  oo 

Thurstbn's  Materials  of  Engineering 3  vols.,  Svo  8  oo 

Part   II. — Iron  and  Steel Svo,  3  50 

Part  HI. — A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents Svo,  2  50 

Tezt-book  of  the  Materials  of  Construction Svo  5  oo 

Wood's  Treatise  on  the  Resistance  of  Materials  and  an  Appendix  on  the 

Preservation  of  Timber Svo,  2  oo 

Elements  of  Analytical  Mechanics Svo,  3  oo 


STEAM-ENGINES  AND  BOILERS. 

Carnot's  Reflections  on  the  Motive  Power  of  Heat.     (Thurston.) i2mo,    i  50 

Dawson's  "Engineering"  and  Electric  Traction  Pocket-book.  .i6mo,  mor.,    4  oo 

Ford's  Boiler  Making  for  Boiler  Makers iSmo,    i  oo 

Goss's  Locomotive  Sparks 8vo,    2  oo 

He  men  way's  Indicator  Practice  and  Steam-engine  Economy 12  mo,    2  oo 

Button's  Mechanical  Engineering  of  Power  Plants Svo,    5  oo 

Heat  and  Heat-engines Svo,    5  oo 

Kent's  Steam-boiler  Economy Svo,   4  oo 

Kneass's  Practice  and  Theory  of  the  Injector Svo.    x  50 

MacCord's  Slide-valves Svo,    2  oo 

Meyer's  Modern  Locomotive  Construction 4to,    10  oo 

Peabody's  Manual  of  the  Steam-engine  Indicator i2mo,    i  50 

Tables  of  the  Properties  of  Saturated  Steam  and  Other  Vapors Svo,    i  oo 

Thermodynamics  of  the  Steam-engine  and  Other  Heat-engines Svo,    5  oo 

Valve-gears  for  Steam-engines Svo,    a  50 

Peabody  and  Miller's  Steam-boilers Svo,   4  oo 

Pray*i  Twenty  Years  with  the  Indicator Large  Svo,    2  50 

Pupln's  Thermodynamics  of  Reversible  Cycles  in  Gases  and  Saturated  Vapors. 

(Osterberg. ) I2mo,  i  25 

Reagan's  Locomotives :  Simple,  Compound,  and  Electric i2mo,  2  50 

Rontgen's  Principles  of  Thermodynamics.     (Du  Bois.) Svo,    5  oo 

Sinclair's  Locomotive  Engine  Running  and  Management i2mo,    a  oo 

Smart's  Handbook  of  Engineering  Laboratory  Practice i2mo,    2  50 

Snow's  Steam-boiler  Practice Svo,   3  oo 

12 


Spangler's  Valve-gears 8vo,    2  50 

Notes  on  Thermodynamics I2mo,    I  oo 

Spangler,  Greene,  and  Marshall's  Elements  of  Steam-engineering 8vo,    3  oo 

Thurston's  Handy  Tables '. 8vo,    i   50 

Manual  of  the  Steam-engine 2  vols..  8vo     10  oo 

Part  I.— History,  Structuce,  and  Theory 8vo,    6  oo 

Part  H. — Design,  Construction,  and  Operation 8vo,    6  oo 

Handbook  of  Engine  and  Boiler  Trials,  and  the  Use  of  the  Indicator  and 

the  Prony  Brake 8vo     5  oo 

Stationary  Steam-engines 8vo,    2  50 

Steam-boiler  Explosions  in  Theory  and  in  Practice 12 mo,    i  50 

Manual  of  Steam-boilers ,  Their  Designs,  Construction,  and  Operation .  8vo,    5  oo 

Weisbach's  Heat,  Steam,  a  id  Steam-engines.     (Du  Bois.) 8vo,    5  oo 

Whitham's  Steam-engine  I  -esign 8vo,    5  oo 

Wilson's  Treatise  on  Steam-boilers.     (Flather.) 1 6mo,    2  50 

Wood's  Thermodynamics.  Heat  Motors,  and  Refrigerating  Machines 8vo,    4  oo 


MECHANICS    A.ND    MACHINERY. 

Barr's  Kinematics  ot  machinery 8vo,  2~so 

Bovey's  Strength  of  Materials  and  Theory  of  Structures 8vo,  7  50 

Chase's  The  Art  of  Pattern-making i2mo,  2  50 

Chordal. — Extracts  from  Letters izmo,  2  oo 

Church's  Mechanics  of  Engineering 8vo  6  oo 


Notes  and  Examples  in  Mechanics 8vo. 

Compton's  First  Lessons  in  Metal-working I2mo, 

Compton  and  De  Groodt's  The  Speed  Lathe I2mo, 


oo 
So 
50 
50 
50 


Cromwell's  Treatise  on  Toothed  Gearing i2mo, 

Treatise  on  Belts  and  Pulleys i2mo, 

Dana's  Text-book  of  Elementary  Mechanics  for  the  Use  of  Colleges  and 

Schools i2mo,    i  50 

Dingey's  Machinery  Pattern  Making i2mo,   2  oo 

Dredge's  Record  of  the  Transportation  Exhibits  Building  of  the  World's 

Columbian  Exposition  of  1893 4to,  half  morocco,    5  oo 

Du  Bois's  Elementary  Principles  of  Mechanics : 

Vol.     I. — Kinematics 8vo,    3  50 

Vol.   II. — Statics 8vo,   4  oo 

Vol.  III.— Kinetics 8vo,    3  50 

Mechanics  of  Engineering.    Vol.  I Small  4to,      7  50 

Vol.  n Small  4to,    10  oo 

Durley's  Kinematics  of  Machines 8vo,    4  oo 

Fitzgerald's  Boston  Machinist i6mo,    i  oo 

Flather's  Dynamometers,  and  the  Measurement  of  Power i2mo,    3  oo 

Rope  Driving I2mo,    2  oo 

Goss's  Locomotive  Sparks 8vo,    2  oo 

Hall's  Car  Lubrication i2mo,    i  oo 

Holly's  Art  of  Saw  Filing i8mo         75 

*  Johnson's  Theoretical  Mechanics X2mo,   3  oo 

Statics  by  Graphic  and  Algebraic  Methods 8vo,    2  oo 

Jones's  Machine  Design: 

Part  I. — Kinematics  of  Machinery 8vo,    i  50 

Part  n. — Form,  Strength,  and  Proportions  of  Parts 8vo,    3  oo 

Kerr's  Power  and  Power  Transmission 8vo,    2  oo 

Lanza's  Applied  Mechanics 8vo,    7  50 

MacCord's  Kinematics;  or,  Practical  Mechanism 8vo,   5  oo 

Velocity  Diagrams 8ro,    i  50 

Maurer's  Technical  Mechanics.     (In  preparation.) 

13 


Merriman's  Text-book  on  the  Mechanics  of  Materials 8vo,  4  oo 

*  Michie's  Elements  of  Analytical  Mechanics 8vo,  4  oo 

Reagan's  Locomotives:  Simple,  Compound,  and  Electric 1 2mo,  2  50 

Reid's  Course'in  Mechanical  Drawing 8vo,  2  oo 

Text-book  of^Mechanical  Drawing  and  Elementary  Machine  Design.  .8vo,  3  oo 

Richards's  Compressed  Air i2mo,  i  50 

Robinson's  Principles  of  Mechanism 8vo,  3  oo 

Ryan,  Norris,  and  Hoxie's  Electrical  Machinery.     (In  preparation.) 

Sinclair's  Locomotive-engine  Running  andjManagement i2mo,  2  oo 

Smith's  Press-working  of  Metals 8vo,  3  oo 

H     Materials  of  Machines i2mo,  i  oo 

Spangler,  Greene,  and  Marshall's  Elements  of  Steam-engineering 8vo,  3  oo 

Thurston's  Treatise  on  Friction  and  Lost  Work  in  Machinery  and  Mill 

Work 8vo,  3  oo 

Animal  as  a  Machine  and  Prime  Motor,  and  the  Laws  of  Energetics .  i2mo,  i  oo 

Warren's  Elements  of  Machine  Construction  and  Drawing 8vo,  7  50 

Weisbach's    Kinematics  land    the  Power  of    Transmission.     (Herrmann — 

Klein.) 8vo,  5  oo 

Machinery  of  Transmission  and  Governors.     (Herrmann — Klein. ).8vo,  5  oo 

Wood's  Elements  of  Analytical  Mechanics 8vo,  3  oo 

Principles  of  Elementary  Mechanics i2mo,  i  25 

Turbines 8vo,  2  50 

The  World's  Columbian  Exposition  of  1893 4to,  i  oo 

METALLURGY. 

Egleston's  Metallurgy  of  Silver,  Gold,  and  Mercury: 

VoL  I. — Silver 8vo,  7  50 

Vol.   IL — Gold  and  Mercury 8vo,  7  So 

**  Iles's  Lead-smelting.     (Postage  9  cents  additional.) i2mo,  2  50 

Keep's  Cast  Iron 8vo,  2  50 

Kunhardt's  Practice  of  Ore  Dressing  in  Europe 8vo,  i  50 

Le  Chatelier's  High-temperature  Measurements.  (Boudouard — Burgess.) .  i2mo,  3  oo 

Metcalf's  SteeL     A  Manual  for  Steel-users i2mo,  2  oo 

Smith's  Materials  of  Machines i2mo,  i  oo 

Thurston's  Materials  of  Engineering.     In  Three  Parts 8vo,  8  oo 

Part  II. — Iron  and  Steel 8vo,  3  50 

Part  III. — A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and   their 

Constituents 8vo,  2  50 

Hike's  Modern  Electrolytic  Copper  Refining 8vo,  3  oo 

MINERALOGY. 

Barringer's  Description  of  Minerals  of  Commercial  Value.     Oblong,  morocco,  2  50 

Boyd's  Resources  of  Southwest  Virginia 8vo,  3  oo 

Map  of  Southwest  Virginia Pocket-book  form,  2  oo 

Brush's  Manual  of  Determinative  Mineralogy.     (Penfield.) 8vo,  4  oo 

Chester's  Catalogue  of  Minerals 8vo,  paper,  i  oo 

Cloth,  i  25 

Dictionary  of  the  Names  of  Minerals 8vo,  3  50 

Dana's  System  of  Mineralogy Large  8vo,  half  leather,    12  50 

First  Appendix  to  Dana's  New  "System  of  Mineralogy.". . .  .Large  8vo,  i  oo 

Text-book  of  Mineralogy 8vo,  4  oo 

Minerals  and  How  to  Study  Them . . . , i2mo,  i  50 

Catalogue  of  American  Localities  of  Minerals Large  8vo,  i  oo 

Manual  of  Mineralogy  and  Petrography i2mo,  2  oo 

Egleston's  Catalogue  of  Minerals  and  Synonyms 8vo,  2  50 

Hussak's  The  Determination  of  Rock-forming  Minerals.     (Smith.)  Small  8vo,  2  oo 

14 


*  Penfield's  Notes  on  Determinative  Mineralogy  and  Record  of  Mineral  Tests. 

8vo,  paper,  o  50 
Rosenbusch's   Microscopical   Physiography   of   the    Rock-making   Minerals. 

(Iddings.) .» 8vo,  5  oo 

*  Tillman's  Text-book  of  Important  Minerals  and  Docks 8vo,  2  oo 

Williams's  Manual  of  Lithology 8vo,  3  oo 

MINING. 

Beard's  Ventilation  of  Mines lamo,  2  50 

Boyd's  Resources  of  Southwest  Virginia 8vo,  3  oo 

Map  of  Southwest  Virginia Pocket-book  form,  2  oo 

*  Drinker's  Tunneling,  Explosive  Compounds,  and  Rock  Drills. 

4to,  half  morocco,  25  oo 

Eissler's  Modern  High  Explosives 8vo,  4  oo 

Fowler's  Sewage  Works  Analyses i2mo,  2  oo 

Goodyear's  Coal-mines  of  the  Western  Coast  of  the  United  States 12 mo,  2  50 

Ihlseng's  Manual  of  Mining  „      8vo,  4  oo 

**  Iles's  Lead-smelting.     (Postage  gc.  additional.) 1 2mo,  2  50 

Kunhardt's  Practice  of  Ore  Dressing  in  Europe 8vo,  i  50 

O'Driscoll's  Notes  on  the  Treatment  of  Gold  Ores 8vo,  2  oo 

*  Walke's  Lectures  on  Explosives 8vo,  4  oo 

Wilson's  Cyanide  Processes I2mo,  i  50 

Chlorination  Process lamo,  i  50 

Hydraulic  and  Placer  Mining i2mo,  2  oo 

Treatise  on  Practical  and  Theoretical  Mine  Ventilation I2mo,  i  25 

SANITARY  SCIENCE. 

Copeland's  Manual  of  Bacteriology.     (In  preparation.) 

Folwell's  Sewerage.     (Designing,  Construction,  and  Maintenance.; 8vo,  3  oo 

Water-supply  Engineering 8vo,  4  oo 

Fuertes's  Water  and  Public  Health i2mo,  i  50 

Water-filtration   Works I2mo,  2  50 

Gerhard's  Guide  to  Sanitary  House-inspection i6mo,  i  oo 

Goodrich's  Economical  Disposal  of  Town's  Refuse Demy  8vo,  3  50 

Hazen's  Filtration  of  Public  Water-supplies 8vo,  3  oo 

Kiersted's  Sewage  Disposal i2mo,  i  25 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State 

Control.     (In  preparation.) 

Mason's   Water-supply.     (Considered   Principally   from   a   Sanitary   Stand- 
point.)   3d  Edition,  Rewritten 8vo,  4  oo 

Examination  of  Water.     (Chemical  and  Bacteriological.) i2mo,  i  25 

Merriman's  Elements  of  Sanitary  Engineering 8vo,  2  oo 

Nichols's  Water-supply.     (Considered  Mainly  from  a  Chemical  and  Sanitary 

Standpoint.)     (1883.) 8vo,  2  50 

Ogden's  Sewer  Design i2mo,  2  oo 

*  Price's  Handbook  on  Sanitation 1 2mo,  i  50 

Richards's  Cost  of  Food.     A  Study  in  Dietaries i2mo,  i  oo 

Cost  of  Living  as  Modified  by  Sanitary^Science I2mo,  I  oo 

Richards  and  Woodman's  Air,  Water,  and  Food  from  a  Sanitary  Stand- 
point  8vo,  2  oo 

*  Richards  and  Williams's  The  DietarylComputer 8vo,  i  50 

Rideal's  Sewage  and  Bacterial  Purification  of  Sewage 8vo,  3  SO 

Turneaure  and  Russell's  Public  Water-supplies 8vo,  5  oo 

Whipple's  Microscopy  of  Drinking-water 8vo,  3  50 

Woodhull's  Notes'and  Military  Hygiene i6mo,  i  50 

15 


MISCELLANEOUS. 

Barker's  Deep-sea  Soundings 8vo,  2  oo 

Emmons's  Geological  Guide-book  of  the  Rocky  Mountain  Excursion  of  the 

International  Congress  of  Geologists Large  8vo,  i  50 

Ferrel's  Popular  Treatise  on  the  Winds 8vo,  4  oo 

Haines's  American  Railway  Management i2mo,  2  50 

Mott's  Composition.'Digestibility ,  and  Nutritive  Value  of  Food.   Mounted  chart,  i  25 

Fallacy  of  the  Present  Theory  of  Sound i6mo,  i  oo 

Ricketts's  History  of  Rensselaer  Polytechnic  Institute,  1824-1894.  Small  8vo,  3  oo 

Rotherham's  Empaasized  New  Testament Large  8vo,  2  oo 

Steel's  Treatise  on  the  Diseases  of  the  Dog 8vo,  3  50 

Totten's  Important  Question  in  Metrology 8vo,  2  50 

The  World's  Columbian  Exposition  of  1893 4to,  i  oo 

Worcester  and  Atkinson.     Small  Hospitals,  Establishment  and  Maintenance, 
and  Suggestions  for  Hospital  Architecture,  with  Plans  for  a  Small 

Hospital i2mo,  i  25 

HEBREW  AND  CHALDEE  TEXT-BOOKS. 

Green's  Grammar  of  the  Hebrew  Language 8vo,  3  oo 

Elementary  Hebrew  Grammar i2mo,  i  25 

Hebrew  Chrestomathy 8vo,  2  oo 

Gesenius's  Hebrew  and  Chaldee  Lexicon  to  the  Old  Testament  Scriptures. 

(Tregelles.) Small  4to,  half  morocco,  5  oo 

Letteris's  Hebrew  Bible 8vo,  2  25 

16 


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